The Argument Principle
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The Argument Principle - Geometric Meaning
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Rouche s Theorem
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Hurwitz s Theorem
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Existence of Logarithms and Roots of Functions
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[{"Name":"The Argument Principle","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Argument Principle (with proof)","Duration":"7m 2s","ChapterTopicVideoID":22906,"CourseChapterTopicPlaylistID":102290,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.750","Text":"In this clip, we\u0027ll talk about the argument principle."},{"Start":"00:03.750 ","End":"00:06.615","Text":"It\u0027s a theorem that is often useful."},{"Start":"00:06.615 ","End":"00:11.655","Text":"We let f be analytic inside and on a closed contour Gamma,"},{"Start":"00:11.655 ","End":"00:17.250","Text":"except for a finite number of poles and we suppose that on Gamma itself,"},{"Start":"00:17.250 ","End":"00:19.935","Text":"there are no zeros and no poles."},{"Start":"00:19.935 ","End":"00:21.960","Text":"Here\u0027s a little picture I found on the web."},{"Start":"00:21.960 ","End":"00:24.330","Text":"This is Gamma, the contour,"},{"Start":"00:24.330 ","End":"00:27.134","Text":"and here are zeros and poles."},{"Start":"00:27.134 ","End":"00:31.800","Text":"What the theorem says is that the integral along Gamma"},{"Start":"00:31.800 ","End":"00:36.675","Text":"of f\u0027 over f is 2Pi i times N minus P,"},{"Start":"00:36.675 ","End":"00:41.240","Text":"where N is the number of zeros inside Gamma and"},{"Start":"00:41.240 ","End":"00:45.860","Text":"P is the number of poles inside Gamma and very important,"},{"Start":"00:45.860 ","End":"00:50.230","Text":"you count each 0 on each pole as many times as its order."},{"Start":"00:50.230 ","End":"00:55.595","Text":"A pole of order 3 is counted like 3 poles or 0 of order 2 like 2 zeros and so on."},{"Start":"00:55.595 ","End":"00:57.590","Text":"There is an alternate form almost the same."},{"Start":"00:57.590 ","End":"01:03.670","Text":"Sometimes we prefer to write the 2Pi i on the left and like this way, no big difference."},{"Start":"01:03.670 ","End":"01:07.415","Text":"You might be wondering why it\u0027s called the argument principle."},{"Start":"01:07.415 ","End":"01:08.960","Text":"I won\u0027t go into it in great detail,"},{"Start":"01:08.960 ","End":"01:15.349","Text":"but basically this integral is equal to i times the increase of the argument"},{"Start":"01:15.349 ","End":"01:17.840","Text":"of f when traveling once around"},{"Start":"01:17.840 ","End":"01:22.235","Text":"C. I\u0027ll say more about this geometric meaning in the next section."},{"Start":"01:22.235 ","End":"01:25.280","Text":"Anyway, let\u0027s get onto the proof."},{"Start":"01:25.280 ","End":"01:30.155","Text":"Notice that the singularities of f\u0027(f),"},{"Start":"01:30.155 ","End":"01:35.420","Text":"the poles and zeros of f. If we have a singularity of f,"},{"Start":"01:35.420 ","End":"01:41.030","Text":"it\u0027s also a singularity of f\u0027 and of the quotient and if we have a 0 of f,"},{"Start":"01:41.030 ","End":"01:43.750","Text":"it\u0027s also because we have a denominator 0."},{"Start":"01:43.750 ","End":"01:49.265","Text":"We have to consider the zeros and the poles of f. Now suppose the zeros,"},{"Start":"01:49.265 ","End":"01:52.040","Text":"enumerate them z_1, z_n,"},{"Start":"01:52.040 ","End":"01:58.240","Text":"and let their orders be m_1 and so on up to m_n respectively."},{"Start":"01:58.240 ","End":"02:02.910","Text":"The poles, we\u0027ll use the letter w, w_1 to w_s."},{"Start":"02:02.910 ","End":"02:07.935","Text":"Let\u0027s say their orders as poles are p_1 through p_s."},{"Start":"02:07.935 ","End":"02:10.380","Text":"By the residue theorem,"},{"Start":"02:10.380 ","End":"02:15.320","Text":"1 over 2Pi i times the integral of f\u0027 over f is equal to the sum of"},{"Start":"02:15.320 ","End":"02:20.870","Text":"the residues of f\u0027 over f at the singularities inside Gamma,"},{"Start":"02:20.870 ","End":"02:24.380","Text":"but these are just the zeros and poles of f."},{"Start":"02:24.380 ","End":"02:30.110","Text":"The sum is over the z_k and over the w_k and then added together."},{"Start":"02:30.110 ","End":"02:37.055","Text":"The plan is to show that each one of these is equal to the order of the"},{"Start":"02:37.055 ","End":"02:44.780","Text":"0 and each one of these is equal to minus the order of the pole."},{"Start":"02:44.780 ","End":"02:46.550","Text":"Then putting that together,"},{"Start":"02:46.550 ","End":"02:49.790","Text":"what we\u0027ll get is that this is equal to the sum of"},{"Start":"02:49.790 ","End":"02:53.490","Text":"the orders of the zeros and that is what we called N,"},{"Start":"02:53.490 ","End":"02:57.110","Text":"it\u0027s counting the zeros with their orders minus this sum,"},{"Start":"02:57.110 ","End":"02:59.855","Text":"which is the sum of the orders of the pole,"},{"Start":"02:59.855 ","End":"03:02.995","Text":"which is the number of poles if you count their order."},{"Start":"03:02.995 ","End":"03:05.390","Text":"It remains to prove 2 things,"},{"Start":"03:05.390 ","End":"03:07.880","Text":"this and this, and then we\u0027re done."},{"Start":"03:07.880 ","End":"03:10.490","Text":"We\u0027ll do the zeros first and then the pole,"},{"Start":"03:10.490 ","End":"03:12.965","Text":"so we have to prove this equality."},{"Start":"03:12.965 ","End":"03:18.590","Text":"We can write f as z minus z_k to the power"},{"Start":"03:18.590 ","End":"03:25.460","Text":"of the order of the 0 times a function which is not 0, z_k."},{"Start":"03:25.460 ","End":"03:29.645","Text":"That\u0027s definition of the order of 0 and f\u0027."},{"Start":"03:29.645 ","End":"03:34.625","Text":"By differentiating this using the product rule gives us this."},{"Start":"03:34.625 ","End":"03:38.990","Text":"Now we divide them f\u0027 over f and we get this."},{"Start":"03:38.990 ","End":"03:47.445","Text":"Then we can cancel by z minus z_k^m_k minus 1 top and bottom, we get this."},{"Start":"03:47.445 ","End":"03:50.115","Text":"We also push the g_k upstairs."},{"Start":"03:50.115 ","End":"03:55.890","Text":"It will be easier. We\u0027ll call the numerator big F and the denominator big G,"},{"Start":"03:55.890 ","End":"03:57.680","Text":"you probably you can guess where I\u0027m going."},{"Start":"03:57.680 ","End":"04:02.060","Text":"It\u0027s going to be a simple pole and we know the formula for the residue."},{"Start":"04:02.060 ","End":"04:03.485","Text":"Anyway, z_k,"},{"Start":"04:03.485 ","End":"04:06.530","Text":"I claim is a 0 of f,"},{"Start":"04:06.530 ","End":"04:11.595","Text":"but not of g. Big F(z_k) is this."},{"Start":"04:11.595 ","End":"04:13.920","Text":"Just plug-in z_k instead of z,"},{"Start":"04:13.920 ","End":"04:15.585","Text":"so we\u0027ve got 0 here,"},{"Start":"04:15.585 ","End":"04:21.645","Text":"and then the g_k cancels and we\u0027re left with just m_k."},{"Start":"04:21.645 ","End":"04:24.795","Text":"This is not 0. On the other hand,"},{"Start":"04:24.795 ","End":"04:29.995","Text":"z_k is a simple 0 of g. We can see that down here, it\u0027s obvious."},{"Start":"04:29.995 ","End":"04:33.080","Text":"We can use the formula for the residue."},{"Start":"04:33.080 ","End":"04:35.255","Text":"We want the residue of f\u0027 over f,"},{"Start":"04:35.255 ","End":"04:42.485","Text":"but that\u0027s equal to the residue of f over g. By the formula for residue at a simple pole,"},{"Start":"04:42.485 ","End":"04:44.450","Text":"we leave the numerator as is,"},{"Start":"04:44.450 ","End":"04:47.855","Text":"differentiate the denominator and plug-in the point."},{"Start":"04:47.855 ","End":"04:49.950","Text":"The numerator, we get F(z_k),"},{"Start":"04:49.950 ","End":"04:51.605","Text":"the derivative of g,"},{"Start":"04:51.605 ","End":"04:55.655","Text":"the derivative of z minus z_k is just 1."},{"Start":"04:55.655 ","End":"04:58.490","Text":"It\u0027s just F(z_k)."},{"Start":"04:58.490 ","End":"05:00.200","Text":"We\u0027ve already computed that here,"},{"Start":"05:00.200 ","End":"05:02.120","Text":"it\u0027s equal to m_k."},{"Start":"05:02.120 ","End":"05:06.565","Text":"That proves the first of the 2 formulas."},{"Start":"05:06.565 ","End":"05:10.320","Text":"Now onto the second with the poles"},{"Start":"05:10.320 ","End":"05:14.145","Text":"instead of zeros and here we get a minus. That\u0027s the difference."},{"Start":"05:14.145 ","End":"05:17.320","Text":"By the definition of pole of order or something,"},{"Start":"05:17.320 ","End":"05:20.725","Text":"in this case p_k, w_k is the pole of order p_k."},{"Start":"05:20.725 ","End":"05:25.055","Text":"We can write it as some function of z over z minus w_k^p_k,"},{"Start":"05:25.055 ","End":"05:29.530","Text":"where this function h_k is not 0 at z_k."},{"Start":"05:29.530 ","End":"05:32.840","Text":"Before differentiating, it\u0027s easier to write"},{"Start":"05:32.840 ","End":"05:36.730","Text":"it with a negative exponent than have a quotient."},{"Start":"05:36.730 ","End":"05:40.010","Text":"We get this. It\u0027s pretty straightforward."},{"Start":"05:40.010 ","End":"05:43.705","Text":"Then divide f\u0027 over f and we get this."},{"Start":"05:43.705 ","End":"05:45.170","Text":"Then we want to simplify."},{"Start":"05:45.170 ","End":"05:49.115","Text":"We want to divide top and bottom by the highest power of p_k."},{"Start":"05:49.115 ","End":"05:52.805","Text":"We can and also push the h_k upstairs."},{"Start":"05:52.805 ","End":"05:57.740","Text":"Just like before, we get this expression and then we\u0027re going to label"},{"Start":"05:57.740 ","End":"06:03.800","Text":"the numerator f and the denominator g. If we plug in w_k,"},{"Start":"06:03.800 ","End":"06:06.890","Text":"then we get w_k minus w_k here."},{"Start":"06:06.890 ","End":"06:10.570","Text":"This part is 0, then the h_k cancels,"},{"Start":"06:10.570 ","End":"06:14.160","Text":"and so that is equal to minus p_k,"},{"Start":"06:14.160 ","End":"06:16.470","Text":"which is not 0."},{"Start":"06:16.470 ","End":"06:20.130","Text":"That\u0027s the f. As of g,"},{"Start":"06:20.130 ","End":"06:24.240","Text":"here is g, it\u0027s a simple 0 of g,"},{"Start":"06:24.240 ","End":"06:26.610","Text":"so not a 0 the numerator,"},{"Start":"06:26.610 ","End":"06:30.580","Text":"simple 0 of denominator so a simple pole,"},{"Start":"06:30.920 ","End":"06:35.675","Text":"big F over G. f\u0027 over f is f over g and we\u0027ve showed that"},{"Start":"06:35.675 ","End":"06:40.929","Text":"the conditions for applying the shortcut formula for simple pole apply."},{"Start":"06:40.929 ","End":"06:42.685","Text":"This is equal to,"},{"Start":"06:42.685 ","End":"06:45.140","Text":"differentiate the denominator, leave the numerator,"},{"Start":"06:45.140 ","End":"06:48.800","Text":"plug-in the point and that comes out to be,"},{"Start":"06:48.800 ","End":"06:52.460","Text":"we already have f of w_k."},{"Start":"06:52.460 ","End":"06:55.890","Text":"That\u0027s here, that\u0027s minus p_k."},{"Start":"06:56.270 ","End":"07:02.590","Text":"This equals minus p_k and that\u0027s what we had to show. We\u0027re done."}],"ID":23757},{"Watched":false,"Name":"Exercise 1","Duration":"1m 23s","ChapterTopicVideoID":22907,"CourseChapterTopicPlaylistID":102290,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.560","Text":"In this exercise, we\u0027ll use the argument principle to compute the integral of"},{"Start":"00:04.560 ","End":"00:10.710","Text":"2z over z^2 plus 1 on the circle of radius 2 circles around 0."},{"Start":"00:10.710 ","End":"00:17.889","Text":"Here\u0027s a reminder of what the argument principle says and here is the picture."},{"Start":"00:17.889 ","End":"00:19.940","Text":"Now if we look at this form,"},{"Start":"00:19.940 ","End":"00:23.360","Text":"f’/f, and we looked at this 2z/z^2 plus 1."},{"Start":"00:23.360 ","End":"00:26.810","Text":"We can see that if we let f(z) be z^2 plus 1,"},{"Start":"00:26.810 ","End":"00:30.255","Text":"then f’ is 2z and it\u0027s exactly in this form."},{"Start":"00:30.255 ","End":"00:32.040","Text":"Let\u0027s look at f now,"},{"Start":"00:32.040 ","End":"00:34.425","Text":"let\u0027s see what its zeros and poles are."},{"Start":"00:34.425 ","End":"00:36.480","Text":"It has no poles, of course,"},{"Start":"00:36.480 ","End":"00:40.950","Text":"z^2 plus 1 is an entire function and the zeros are when z^2 plus 1 is 0,"},{"Start":"00:40.950 ","End":"00:43.480","Text":"so z is plus or minus i."},{"Start":"00:43.480 ","End":"00:47.570","Text":"Now, here\u0027s the circle of radius 2,"},{"Start":"00:47.570 ","End":"00:50.300","Text":"f is analytic on the circle and inside"},{"Start":"00:50.300 ","End":"00:53.750","Text":"the circle and it has no zeros or poles on the circle,"},{"Start":"00:53.750 ","End":"00:56.835","Text":"so we can apply the principle,"},{"Start":"00:56.835 ","End":"01:03.600","Text":"we need n and p. It\u0027s fairly clear that these two zeros are simple zeros,"},{"Start":"01:03.600 ","End":"01:06.150","Text":"so n=2 and there are no poles,"},{"Start":"01:06.150 ","End":"01:08.070","Text":"so p is 0."},{"Start":"01:08.070 ","End":"01:10.550","Text":"Using the argument principle,"},{"Start":"01:10.550 ","End":"01:13.745","Text":"this integral is f’/f,"},{"Start":"01:13.745 ","End":"01:18.905","Text":"which is 2Pi i times n minus p,"},{"Start":"01:18.905 ","End":"01:23.670","Text":"which is just 4Pi i and we\u0027re done."}],"ID":23758},{"Watched":false,"Name":"Exercise 2","Duration":"1m 42s","ChapterTopicVideoID":22908,"CourseChapterTopicPlaylistID":102290,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.275","Text":"In this exercise, we\u0027re given this function f(z),"},{"Start":"00:04.275 ","End":"00:09.135","Text":"and we want to compute the integral of f\u0027 over f using the argument principle."},{"Start":"00:09.135 ","End":"00:14.625","Text":"The integral is on the circle centered at 0 with radius 2."},{"Start":"00:14.625 ","End":"00:17.595","Text":"This is the formula for the argument principle,"},{"Start":"00:17.595 ","End":"00:22.560","Text":"and if you\u0027d like it, here\u0027s a diagram."},{"Start":"00:22.560 ","End":"00:31.845","Text":"Let\u0027s check what the 0s and poles are of f. There\u0027s only 1 pole, 0,"},{"Start":"00:31.845 ","End":"00:35.044","Text":"and its order is 2,"},{"Start":"00:35.044 ","End":"00:36.675","Text":"because of the 2 here,"},{"Start":"00:36.675 ","End":"00:40.275","Text":"and 0 is not a 0 of the numerator,"},{"Start":"00:40.275 ","End":"00:42.875","Text":"and it\u0027s inside the circle."},{"Start":"00:42.875 ","End":"00:45.245","Text":"You know what, let\u0027s put the picture already."},{"Start":"00:45.245 ","End":"00:49.700","Text":"The 0s are minus 1 and minus 3,"},{"Start":"00:49.700 ","End":"00:51.362","Text":"they\u0027re both of order 1."},{"Start":"00:51.362 ","End":"00:58.670","Text":"But the only 1 that\u0027s inside the circle is z equals minus 1."},{"Start":"00:58.670 ","End":"01:02.690","Text":"Note that all the conditions are met that the function"},{"Start":"01:02.690 ","End":"01:06.637","Text":"f is analytic in the disc of radius 2,"},{"Start":"01:06.637 ","End":"01:08.611","Text":"except for a finite number of 0s and poles,"},{"Start":"01:08.611 ","End":"01:11.870","Text":"and none of these is on the circumference,"},{"Start":"01:11.870 ","End":"01:13.459","Text":"on the contour itself."},{"Start":"01:13.459 ","End":"01:15.770","Text":"So we can apply the theorem,"},{"Start":"01:15.770 ","End":"01:18.490","Text":"so n is equal to 1,"},{"Start":"01:18.490 ","End":"01:24.610","Text":"because there\u0027s one 0 of order 1 inside the contour, and p=2."},{"Start":"01:24.610 ","End":"01:26.210","Text":"There\u0027s only 1 pole,"},{"Start":"01:26.210 ","End":"01:29.780","Text":"but it has order 2, so it counts double."},{"Start":"01:29.780 ","End":"01:37.310","Text":"So the formula says that this integral is equal to 2Pii times n minus p. N is 1, P is 2."},{"Start":"01:37.310 ","End":"01:42.690","Text":"So this just comes out to be minus 2Pii, and we\u0027re done."}],"ID":23759},{"Watched":false,"Name":"Exercise 3","Duration":"3m 36s","ChapterTopicVideoID":22909,"CourseChapterTopicPlaylistID":102290,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.280","Text":"In this exercise, we\u0027re going to use the argument principle to compute this integral."},{"Start":"00:05.280 ","End":"00:08.880","Text":"The integral is on the unit circle,"},{"Start":"00:08.880 ","End":"00:15.000","Text":"and it\u0027s 3 sine 2z over sine^2z minus 0.5."},{"Start":"00:15.000 ","End":"00:22.965","Text":"This is the formulation of the principal and here\u0027s a diagram, if it helps."},{"Start":"00:22.965 ","End":"00:28.920","Text":"Let\u0027s let f(z) be the denominator here and see if that does us any good."},{"Start":"00:28.920 ","End":"00:33.420","Text":"Mentally, I\u0027m thinking that the derivative of sine^2z is 2 sine z,"},{"Start":"00:33.420 ","End":"00:35.430","Text":"cosine z, which is sine 2z."},{"Start":"00:35.430 ","End":"00:37.200","Text":"Let\u0027s see what happens."},{"Start":"00:37.200 ","End":"00:39.240","Text":"F\u0027 like I said,"},{"Start":"00:39.240 ","End":"00:41.955","Text":"is 2 sine z cosine z."},{"Start":"00:41.955 ","End":"00:47.900","Text":"This is equal to by famous trigonometric formula for sine of a double angle,"},{"Start":"00:47.900 ","End":"00:50.060","Text":"it\u0027s assigned to z."},{"Start":"00:50.060 ","End":"00:52.510","Text":"It\u0027s very similar to what we have here."},{"Start":"00:52.510 ","End":"00:58.940","Text":"Our integral for just bring the 3 out in front will be f\u0027 over f. In a moment,"},{"Start":"00:58.940 ","End":"01:03.065","Text":"we\u0027ll check that the conditions are satisfied for applying the argument principle."},{"Start":"01:03.065 ","End":"01:07.880","Text":"But if they do, this will be equal to 2Pi times n minus p. Now we need to"},{"Start":"01:07.880 ","End":"01:15.140","Text":"find the 0s in poles of f(z) inside the contour."},{"Start":"01:15.140 ","End":"01:23.315","Text":"f has no poles at all because this sine^2z minus 0.5 is entire."},{"Start":"01:23.315 ","End":"01:27.740","Text":"We need to find the 0s of f inside the disk,"},{"Start":"01:27.740 ","End":"01:32.780","Text":"and then we\u0027ll check to see that none of them are actually on absolute value z=1."},{"Start":"01:32.780 ","End":"01:35.405","Text":"That is only a finite number inside."},{"Start":"01:35.405 ","End":"01:38.780","Text":"This is the equation we have."},{"Start":"01:38.780 ","End":"01:41.515","Text":"Sine^2z equals 1.5."},{"Start":"01:41.515 ","End":"01:43.300","Text":"Now, many ways to solve this."},{"Start":"01:43.300 ","End":"01:47.525","Text":"I\u0027m going to use the formula for cosine 2z."},{"Start":"01:47.525 ","End":"01:50.870","Text":"Then I won\u0027t have to deal with square root plus or minus."},{"Start":"01:50.870 ","End":"01:54.230","Text":"Cosine 2z is 1 minus 2 sine^2z."},{"Start":"01:54.230 ","End":"01:59.990","Text":"This is 0.5 times 2 is 1.1 minus 1 is 0."},{"Start":"01:59.990 ","End":"02:03.785","Text":"We get the cosine 2z is 0."},{"Start":"02:03.785 ","End":"02:09.290","Text":"That means that 2z is Pi over 2 plus"},{"Start":"02:09.290 ","End":"02:14.325","Text":"multiples of Pi is like 90 degrees or 270 degrees and so on."},{"Start":"02:14.325 ","End":"02:16.160","Text":"It\u0027s where the cosine is 0."},{"Start":"02:16.160 ","End":"02:22.150","Text":"Dividing by 2, z is Pi over 4 plus multiples of Pi over 2."},{"Start":"02:22.150 ","End":"02:24.800","Text":"If we put some values of n in,"},{"Start":"02:24.800 ","End":"02:28.130","Text":"let\u0027s say n is 0,"},{"Start":"02:28.130 ","End":"02:31.010","Text":"is here and is 1 and it\u0027s 2 and it\u0027s minus 1,"},{"Start":"02:31.010 ","End":"02:33.530","Text":"minus 2, you can see some of the values."},{"Start":"02:33.530 ","End":"02:38.865","Text":"The only ones inside the unit circle would be these 2."},{"Start":"02:38.865 ","End":"02:40.515","Text":"Pi is roughly 3,"},{"Start":"02:40.515 ","End":"02:41.865","Text":"certainly less than 4."},{"Start":"02:41.865 ","End":"02:46.755","Text":"Pi over 4, minus Pi over 4 between 1 and minus 1."},{"Start":"02:46.755 ","End":"02:49.725","Text":"Everything else goes outside."},{"Start":"02:49.725 ","End":"02:54.687","Text":"There\u0027s 2 which is a finite number inside and none of these are actually on z=1."},{"Start":"02:54.687 ","End":"02:58.570","Text":"There are simple 0s of order 1."},{"Start":"02:58.570 ","End":"03:02.195","Text":"We have to show that the derivative is not 0."},{"Start":"03:02.195 ","End":"03:05.630","Text":"We just substitute z plus or minus Pi over 4."},{"Start":"03:05.630 ","End":"03:08.015","Text":"It\u0027s sine of plus or minus Pi over 2,"},{"Start":"03:08.015 ","End":"03:10.120","Text":"just plus or minus 1."},{"Start":"03:10.120 ","End":"03:11.900","Text":"Then 2 and 1 here."},{"Start":"03:11.900 ","End":"03:14.574","Text":"Anyway, either way it\u0027s not 0."},{"Start":"03:14.574 ","End":"03:17.305","Text":"That means that n equals 2,"},{"Start":"03:17.305 ","End":"03:20.150","Text":"because we have an order 1 plus an order of 1."},{"Start":"03:20.150 ","End":"03:23.840","Text":"1 plus 1 is 2 and that there are no poles, so p is 0."},{"Start":"03:23.840 ","End":"03:27.830","Text":"This integral which is 3 times 2 Pi n minus p,"},{"Start":"03:27.830 ","End":"03:31.800","Text":"n minus p comes out to be 2."},{"Start":"03:31.800 ","End":"03:34.005","Text":"2 times 2 times 3 is 12."},{"Start":"03:34.005 ","End":"03:37.810","Text":"We have 12Pii and we\u0027re done."}],"ID":23760},{"Watched":false,"Name":"Exercise 4","Duration":"3m 55s","ChapterTopicVideoID":22910,"CourseChapterTopicPlaylistID":102290,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.525","Text":"In this exercise, we have a function f(z),"},{"Start":"00:03.525 ","End":"00:06.810","Text":"which is equal to this."},{"Start":"00:06.810 ","End":"00:12.690","Text":"The idea here is to compute this integral using the argument principle."},{"Start":"00:12.690 ","End":"00:14.730","Text":"But it\u0027s broken up into 2 steps."},{"Start":"00:14.730 ","End":"00:17.850","Text":"First, we find the 0s and poles of F together with"},{"Start":"00:17.850 ","End":"00:22.650","Text":"their orders that are in the domain absolute value of z less than 4."},{"Start":"00:22.650 ","End":"00:27.990","Text":"In other words, in the disk centered at 0 and radius 4."},{"Start":"00:27.990 ","End":"00:31.590","Text":"Here\u0027s a reminder of the argument principle,"},{"Start":"00:31.590 ","End":"00:33.675","Text":"at least the basic formula in it."},{"Start":"00:33.675 ","End":"00:36.270","Text":"We\u0027ll start with part a."},{"Start":"00:36.270 ","End":"00:41.885","Text":"The only 0 of f(z) is where z is equal to 2,"},{"Start":"00:41.885 ","End":"00:44.540","Text":"because that\u0027s the only place the numerator can be 0."},{"Start":"00:44.540 ","End":"00:50.255","Text":"It\u0027s clear, and it\u0027s of order 2 because of the squared here and 2 is,"},{"Start":"00:50.255 ","End":"00:53.075","Text":"of course, inside this circle."},{"Start":"00:53.075 ","End":"00:58.760","Text":"The poles of f occur when the denominator is 0,"},{"Start":"00:58.760 ","End":"01:04.660","Text":"meaning that either e to the 2z minus 1 is 0 or z is 0."},{"Start":"01:04.660 ","End":"01:08.450","Text":"It\u0027s going to turn out that z equals 0 is also a 0 of this,"},{"Start":"01:08.450 ","End":"01:12.830","Text":"so that will boost the order of 0 and we\u0027ll see."},{"Start":"01:12.830 ","End":"01:14.600","Text":"We need to solve this equation."},{"Start":"01:14.600 ","End":"01:17.810","Text":"E to the 2z minus 1 equals 0."},{"Start":"01:17.810 ","End":"01:22.280","Text":"I\u0027ll bring the picture already might help."},{"Start":"01:22.280 ","End":"01:30.800","Text":"That means that e to the 2z equals 1 and the general form of 1 is e to the I 2 Pi k,"},{"Start":"01:30.800 ","End":"01:34.325","Text":"where k is any integer."},{"Start":"01:34.325 ","End":"01:37.830","Text":"2z equals I 2pi k,"},{"Start":"01:37.830 ","End":"01:43.115","Text":"z equals I Pi k. Now there\u0027s infinitely many of these,"},{"Start":"01:43.115 ","End":"01:47.165","Text":"but there\u0027s only finitely many inside the disk."},{"Start":"01:47.165 ","End":"01:50.120","Text":"As we can see, there\u0027s only 3 of them."},{"Start":"01:50.120 ","End":"01:52.160","Text":"When k is 0,"},{"Start":"01:52.160 ","End":"01:53.750","Text":"1 or minus 1,"},{"Start":"01:53.750 ","End":"02:00.000","Text":"then the absolute value of the modulus of this one is 0 and this one is Pi,"},{"Start":"02:00.000 ","End":"02:04.050","Text":"which is between 3 and 4 in any event it\u0027s inside."},{"Start":"02:04.050 ","End":"02:08.480","Text":"But if you take K above 1 or below minus 1,"},{"Start":"02:08.480 ","End":"02:10.760","Text":"it will be outside the circle already."},{"Start":"02:10.760 ","End":"02:14.990","Text":"I guess I should point out that none of these poles or 0s is actually on"},{"Start":"02:14.990 ","End":"02:19.985","Text":"the circle which we had to make sure of in order to use the argument principle."},{"Start":"02:19.985 ","End":"02:27.605","Text":"These are all simple 0s of e to the 2z minus 1 because they\u0027re not 0s of the derivative,"},{"Start":"02:27.605 ","End":"02:28.850","Text":"the derivative is 2,"},{"Start":"02:28.850 ","End":"02:31.405","Text":"e to the 2z and that\u0027s never 0."},{"Start":"02:31.405 ","End":"02:35.405","Text":"Just to summarizing that about the denominator."},{"Start":"02:35.405 ","End":"02:40.880","Text":"These 3 are 0s of order 2 of this e"},{"Start":"02:40.880 ","End":"02:46.552","Text":"to the 2z minus 1^2 because the 2 means that not simple anymore,"},{"Start":"02:46.552 ","End":"02:48.040","Text":"they\u0027re now of order 2,"},{"Start":"02:48.040 ","End":"02:53.160","Text":"and 0 is a 0 of order 3 of z^3."},{"Start":"02:53.160 ","End":"02:55.490","Text":"Of course one of them is a 0 of the numerator because only"},{"Start":"02:55.490 ","End":"02:58.175","Text":"2 is a 0 of the numerator and that\u0027s not here."},{"Start":"02:58.175 ","End":"03:04.095","Text":"Now, 0 from these 3 is a pole of order 3 plus 2 equals 5,"},{"Start":"03:04.095 ","End":"03:08.630","Text":"because we take the 2 from here and the 3 from here."},{"Start":"03:08.630 ","End":"03:14.475","Text":"But the others remain of order 2 the plus or minus Pi I."},{"Start":"03:14.475 ","End":"03:16.740","Text":"Here we have pole of order 2,"},{"Start":"03:16.740 ","End":"03:18.600","Text":"order 5, and order 2."},{"Start":"03:18.600 ","End":"03:21.525","Text":"This is a 0 of order 2."},{"Start":"03:21.525 ","End":"03:23.400","Text":"Now on to part b."},{"Start":"03:23.400 ","End":"03:28.105","Text":"From part a, what we have is that n equals 2,"},{"Start":"03:28.105 ","End":"03:32.999","Text":"that\u0027s the 0 of order 2 and p,"},{"Start":"03:32.999 ","End":"03:36.825","Text":"like we said, is going to be 5 and 5 here and 2 here."},{"Start":"03:36.825 ","End":"03:39.255","Text":"That\u0027s 9 altogether."},{"Start":"03:39.255 ","End":"03:42.975","Text":"N minus p is minus 7."},{"Start":"03:42.975 ","End":"03:50.430","Text":"Then the formula is that 1 over 2 Pi i the integral of f\u0027 over f is n minus p,"},{"Start":"03:50.430 ","End":"03:52.440","Text":"and so it\u0027s minus 7."},{"Start":"03:52.440 ","End":"03:55.360","Text":"That\u0027s the answer and we\u0027re done."}],"ID":23761}],"Thumbnail":null,"ID":102290},{"Name":"The Argument Principle - Geometric Meaning","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Argument Principle - geometric meaning","Duration":"6m 21s","ChapterTopicVideoID":22923,"CourseChapterTopicPlaylistID":102291,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"In the previous clip, we learned about the Argument Principle."},{"Start":"00:03.630 ","End":"00:08.415","Text":"I\u0027d like to show you what is the geometric meaning of the Argument Principle."},{"Start":"00:08.415 ","End":"00:12.945","Text":"Let\u0027s first briefly review the Argument Principle."},{"Start":"00:12.945 ","End":"00:17.565","Text":"We have f which is analytic inside and on a closed contour Gamma,"},{"Start":"00:17.565 ","End":"00:20.295","Text":"except for a finite number of poles."},{"Start":"00:20.295 ","End":"00:22.140","Text":"We suppose that none of these poles,"},{"Start":"00:22.140 ","End":"00:26.820","Text":"also that there are no 0s on the contour Gamma itself."},{"Start":"00:26.820 ","End":"00:31.560","Text":"In that case, we have a formula for the number of 0s minus the number"},{"Start":"00:31.560 ","End":"00:36.113","Text":"of poles which is given by this integral,"},{"Start":"00:36.113 ","End":"00:39.965","Text":"and n is the number of zeros inside Gamma,"},{"Start":"00:39.965 ","End":"00:42.530","Text":"P is the number of poles inside Gamma."},{"Start":"00:42.530 ","End":"00:45.470","Text":"This is some domain D inside Gamma."},{"Start":"00:45.470 ","End":"00:48.830","Text":"But the important thing is to count each 0 and pole"},{"Start":"00:48.830 ","End":"00:53.495","Text":"together with its order or multiplicity in the case of a 0."},{"Start":"00:53.495 ","End":"00:55.325","Text":"That was the brief review."},{"Start":"00:55.325 ","End":"01:01.280","Text":"Now, let\u0027s say that Gamma in parametric form is given by Gamma of t,"},{"Start":"01:01.280 ","End":"01:04.145","Text":"the function from an interval (a,"},{"Start":"01:04.145 ","End":"01:07.275","Text":"b) and you want it to be a closed contour."},{"Start":"01:07.275 ","End":"01:12.195","Text":"Gamma of a equals Gamma of b and typically but not always,"},{"Start":"01:12.195 ","End":"01:15.310","Text":"(a, b) is the interval 0-2Pi."},{"Start":"01:15.310 ","End":"01:19.895","Text":"Let\u0027s see if we can figure out this integral using a substitution."},{"Start":"01:19.895 ","End":"01:24.430","Text":"We\u0027ll let z be Gamma of t, and dz,"},{"Start":"01:24.430 ","End":"01:26.588","Text":"of course, its Gamma prime of (t)dt,"},{"Start":"01:26.588 ","End":"01:29.985","Text":"and t moves from a to b."},{"Start":"01:29.985 ","End":"01:35.630","Text":"That gives us the following as a real integral on the interval (a,"},{"Start":"01:35.630 ","End":"01:38.330","Text":"b) and now I\u0027ll make another substitution."},{"Start":"01:38.330 ","End":"01:43.610","Text":"We\u0027ll let w be f of Gamma of t,"},{"Start":"01:43.610 ","End":"01:50.070","Text":"and then dw will be all of this numerator."},{"Start":"01:50.530 ","End":"01:53.940","Text":"As t goes from a to b,"},{"Start":"01:53.940 ","End":"01:58.700","Text":"Gamma of t goes around the curve Gamma and so w is on"},{"Start":"01:58.700 ","End":"02:03.995","Text":"the image of Gamma under f and so we get this integral,"},{"Start":"02:03.995 ","End":"02:09.170","Text":"just 1 over w dw and the integral is on"},{"Start":"02:09.170 ","End":"02:14.870","Text":"the image of Gamma under f. Before we get to the geometric meaning,"},{"Start":"02:14.870 ","End":"02:16.280","Text":"we need another concept,"},{"Start":"02:16.280 ","End":"02:20.570","Text":"the concept of an index of a curve around the point,"},{"Start":"02:20.570 ","End":"02:22.700","Text":"also known as the winding number."},{"Start":"02:22.700 ","End":"02:25.145","Text":"Let\u0027s say Gamma is a contour."},{"Start":"02:25.145 ","End":"02:30.420","Text":"Then we say that the index of Gamma relative to a point that\u0027s not on Gamma,"},{"Start":"02:30.420 ","End":"02:33.980","Text":"z_0, is given by this integral."},{"Start":"02:33.980 ","End":"02:35.270","Text":"This is a notation."},{"Start":"02:35.270 ","End":"02:37.610","Text":"The winding number or index of Gamma"},{"Start":"02:37.610 ","End":"02:43.885","Text":"around z_0 equals 1/2Pi i integral on Gamma 1 over z minus z_0 dz."},{"Start":"02:43.885 ","End":"02:47.720","Text":"This integral gives us the number of times that"},{"Start":"02:47.720 ","End":"02:52.280","Text":"the contour wraps around the point anticlockwise,"},{"Start":"02:52.280 ","End":"02:54.200","Text":"and it\u0027s always an integer."},{"Start":"02:54.200 ","End":"02:57.410","Text":"If it\u0027s clockwise, then it\u0027s negative."},{"Start":"02:57.410 ","End":"03:01.700","Text":"Here\u0027s an example of a winding number 2."},{"Start":"03:01.700 ","End":"03:03.500","Text":"Let\u0027s say we start here,"},{"Start":"03:03.500 ","End":"03:05.920","Text":"then we go around."},{"Start":"03:05.920 ","End":"03:11.675","Text":"Altogether, we wrap around this point twice."},{"Start":"03:11.675 ","End":"03:15.030","Text":"I have an animation of something similar."},{"Start":"03:21.110 ","End":"03:24.340","Text":"Here it is again."},{"Start":"03:29.690 ","End":"03:33.685","Text":"Here are some examples of winding numbers."},{"Start":"03:33.685 ","End":"03:38.424","Text":"If the curve doesn\u0027t wrap around the point,"},{"Start":"03:38.424 ","End":"03:40.930","Text":"then that\u0027s winding number 0."},{"Start":"03:40.930 ","End":"03:44.230","Text":"Negatives of a counterclockwise once around,"},{"Start":"03:44.230 ","End":"03:47.875","Text":"twice around and here, once around,"},{"Start":"03:47.875 ","End":"03:52.405","Text":"anticlockwise, twice around, just like in this picture almost,"},{"Start":"03:52.405 ","End":"03:55.060","Text":"and then 3 times around and so on."},{"Start":"03:55.060 ","End":"03:59.980","Text":"We can write Gamma in exponential form as r(t)e^i Theta,"},{"Start":"03:59.980 ","End":"04:04.240","Text":"where r and Theta are functions of t and are continuous."},{"Start":"04:04.240 ","End":"04:09.215","Text":"Because it\u0027s a closed contour and Gamma of a equals Gamma of b,"},{"Start":"04:09.215 ","End":"04:17.850","Text":"then r(a)e^i Theta at a is equal to r(b) e^i Theta of b."},{"Start":"04:17.850 ","End":"04:21.065","Text":"So the r\u0027s must be equal,"},{"Start":"04:21.065 ","End":"04:26.360","Text":"but Theta is only determined uniquely up to multiples of 2Pi,"},{"Start":"04:26.360 ","End":"04:33.155","Text":"and this n is actually the winding number around 0."},{"Start":"04:33.155 ","End":"04:34.850","Text":"We take the angle at the start,"},{"Start":"04:34.850 ","End":"04:37.670","Text":"the angle at the end and you subtract and you get"},{"Start":"04:37.670 ","End":"04:41.420","Text":"a multiple of 2Pi and that multiple is the winding number."},{"Start":"04:41.420 ","End":"04:45.290","Text":"We can actually get a formula for n which is this,"},{"Start":"04:45.290 ","End":"04:47.615","Text":"n of Gamma around 0,"},{"Start":"04:47.615 ","End":"04:52.305","Text":"which is the end angle minus the start angle,"},{"Start":"04:52.305 ","End":"04:56.660","Text":"and divide that by 2Pi to get the number of whole circles."},{"Start":"04:56.660 ","End":"05:00.920","Text":"This is equal to 1/2 Pi and the numerator is"},{"Start":"05:00.920 ","End":"05:05.840","Text":"denoted as the Delta of the argument of Gamma,"},{"Start":"05:05.840 ","End":"05:07.510","Text":"Theta is the argument of Gamma."},{"Start":"05:07.510 ","End":"05:12.495","Text":"So it\u0027s a Delta of Theta as z wraps around Gamma."},{"Start":"05:12.495 ","End":"05:18.290","Text":"The increase in the argument of z as z goes around the curve Gamma."},{"Start":"05:18.290 ","End":"05:23.090","Text":"Earlier, we had this formula and then we\u0027re going to put this together."},{"Start":"05:23.090 ","End":"05:28.235","Text":"Looking back, this was the formula for the winding number."},{"Start":"05:28.235 ","End":"05:29.850","Text":"In our case,"},{"Start":"05:29.850 ","End":"05:32.755","Text":"there is no w minus w_0,"},{"Start":"05:32.755 ","End":"05:35.985","Text":"it\u0027s wrapping around 0."},{"Start":"05:35.985 ","End":"05:39.875","Text":"It\u0027s the winding number of f of Gamma around 0."},{"Start":"05:39.875 ","End":"05:45.045","Text":"But we also have that this is equal to this."},{"Start":"05:45.045 ","End":"05:50.090","Text":"We can take these 2 equalities and say that the integral of f\"/f,"},{"Start":"05:50.090 ","End":"05:53.480","Text":"1/2 Pi i is equal to this expression,"},{"Start":"05:53.480 ","End":"05:59.060","Text":"1/2 Pi times the increase in the argument."},{"Start":"05:59.060 ","End":"06:05.600","Text":"We can also apply the Argument Principle because it\u0027s got the same integral on the left,"},{"Start":"06:05.600 ","End":"06:10.670","Text":"and that\u0027s n-p. We can actually say that combining them all,"},{"Start":"06:10.670 ","End":"06:13.130","Text":"that this is equal to this,"},{"Start":"06:13.130 ","End":"06:18.695","Text":"is equal to this and this is also equal to the winding number."},{"Start":"06:18.695 ","End":"06:22.140","Text":"Okay, that concludes this clip."}],"ID":23762},{"Watched":false,"Name":"Example 1","Duration":"2m 45s","ChapterTopicVideoID":22924,"CourseChapterTopicPlaylistID":102291,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.120","Text":"In this example exercise,"},{"Start":"00:03.120 ","End":"00:09.690","Text":"we have that f(z)=z^2 and Gamma is the unit circle."},{"Start":"00:09.690 ","End":"00:13.530","Text":"This is the standard parameterization of the unit circle."},{"Start":"00:13.530 ","End":"00:17.520","Text":"We have to show that our usual integral,"},{"Start":"00:17.520 ","End":"00:23.460","Text":"f\u0027 over f and the 1 over 2Pi i here is equal to 1 over"},{"Start":"00:23.460 ","End":"00:30.105","Text":"2Pi times the increase in the argument of f(z) as z traverses Gamma,"},{"Start":"00:30.105 ","End":"00:32.235","Text":"that\u0027s what we had in the tutorial."},{"Start":"00:32.235 ","End":"00:37.470","Text":"We have to show that it hold by evaluating each side separately and comparing,"},{"Start":"00:37.470 ","End":"00:40.180","Text":"and here\u0027s a picture of Gamma."},{"Start":"00:40.180 ","End":"00:41.885","Text":"Start with the left-hand side,"},{"Start":"00:41.885 ","End":"00:47.600","Text":"and we will evaluate this using the argument principle, N minus P,"},{"Start":"00:47.600 ","End":"00:51.620","Text":"where N is the number of 0s and P is the number of poles"},{"Start":"00:51.620 ","End":"00:55.985","Text":"of f. We have a 0 of order 2 for f,"},{"Start":"00:55.985 ","End":"00:59.630","Text":"f(z) is z squared is 0 of order 2 and no poles."},{"Start":"00:59.630 ","End":"01:03.530","Text":"The 0 is inside Gamma."},{"Start":"01:03.530 ","End":"01:07.200","Text":"A 0 of order 2 and no poles, N is 2,"},{"Start":"01:07.200 ","End":"01:13.455","Text":"P is 0 and that means that N minus P is 2."},{"Start":"01:13.455 ","End":"01:15.240","Text":"That\u0027s the left-hand side."},{"Start":"01:15.240 ","End":"01:17.730","Text":"Now onto the right-hand side,"},{"Start":"01:17.730 ","End":"01:21.050","Text":"Gamma is equal to this."},{"Start":"01:21.050 ","End":"01:27.374","Text":"This is the parameterization by Theta, the unit circle."},{"Start":"01:27.374 ","End":"01:31.380","Text":"Now if z is e^i Theta, then f(z),"},{"Start":"01:31.380 ","End":"01:35.280","Text":"which is z^2, is e^i2 Theta,"},{"Start":"01:35.280 ","End":"01:41.220","Text":"so that the image of Gamma is the set of all e^i2 Theta,"},{"Start":"01:41.220 ","End":"01:43.735","Text":"where Theta goes from 0 to 2Pi."},{"Start":"01:43.735 ","End":"01:47.050","Text":"If you think about it because of the 2 here,"},{"Start":"01:47.050 ","End":"01:50.899","Text":"it actually encircles twice around."},{"Start":"01:50.899 ","End":"01:53.915","Text":"It stays on the unit circle because r is still 1,"},{"Start":"01:53.915 ","End":"01:56.165","Text":"but it goes around twice."},{"Start":"01:56.165 ","End":"02:02.389","Text":"The argument of f(z) is a 2 Theta here, that\u0027s the argument."},{"Start":"02:02.389 ","End":"02:06.560","Text":"But, plus in principle multiples of 2Pi,"},{"Start":"02:06.560 ","End":"02:10.490","Text":"but we can actually not bother with the 2NPi because it\u0027ll cancel out anyway."},{"Start":"02:10.490 ","End":"02:14.270","Text":"I just put it in just to be a bit pedantic."},{"Start":"02:14.270 ","End":"02:19.685","Text":"If we took 2 Theta plus 2nPi and we substituted once,"},{"Start":"02:19.685 ","End":"02:23.630","Text":"we let Theta equals 0,"},{"Start":"02:23.630 ","End":"02:26.940","Text":"and once Theta equals 2Pi and subtract,"},{"Start":"02:26.940 ","End":"02:28.800","Text":"we\u0027ll get 4Pi,"},{"Start":"02:28.800 ","End":"02:35.525","Text":"the 2NPi won\u0027t contribute the increase in the argument is still 4Pi."},{"Start":"02:35.525 ","End":"02:39.725","Text":"The other side is 4Pi over 2Pi, which is 2,"},{"Start":"02:39.725 ","End":"02:41.915","Text":"and this 2 is equal to this 2,"},{"Start":"02:41.915 ","End":"02:46.620","Text":"and so we have verified the equality and we\u0027re done."}],"ID":23763},{"Watched":false,"Name":"Example 2","Duration":"2m 11s","ChapterTopicVideoID":22925,"CourseChapterTopicPlaylistID":102291,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"In this example, we\u0027re going to compute the following,"},{"Start":"00:03.720 ","End":"00:07.950","Text":"which is 1 over 2Pi times the increase in"},{"Start":"00:07.950 ","End":"00:14.505","Text":"the argument of f(z) as z traverses the contour Gamma."},{"Start":"00:14.505 ","End":"00:18.495","Text":"f(z) is z squared minus z,"},{"Start":"00:18.495 ","End":"00:23.715","Text":"and the contour Gamma is given by this parameterization."},{"Start":"00:23.715 ","End":"00:27.360","Text":"If you think about it, that\u0027s a circle of radius 2."},{"Start":"00:27.360 ","End":"00:31.425","Text":"In fact, it\u0027s a circle centered at 0 with radius 2."},{"Start":"00:31.425 ","End":"00:36.060","Text":"This expression according to the formula we had for"},{"Start":"00:36.060 ","End":"00:41.910","Text":"the geometric interpretation of the argument theorem is equal to this integral."},{"Start":"00:41.910 ","End":"00:44.510","Text":"This is N minus P,"},{"Start":"00:44.510 ","End":"00:45.995","Text":"where as usual,"},{"Start":"00:45.995 ","End":"00:53.120","Text":"N is the number of zeros and P is the number of poles that are inside the contour Gamma,"},{"Start":"00:53.120 ","End":"00:55.775","Text":"in the open unit disc of radius 2."},{"Start":"00:55.775 ","End":"01:02.405","Text":"Now our function has simple zeros at 0 and 1 and they\u0027re both inside Gamma."},{"Start":"01:02.405 ","End":"01:05.545","Text":"N equals 2, 1 and 1,"},{"Start":"01:05.545 ","End":"01:09.679","Text":"and also P equals 0 because it\u0027s a polynomial,"},{"Start":"01:09.679 ","End":"01:15.305","Text":"certainly has no poles anywhere and definitely not inside Gamma or on Gamma."},{"Start":"01:15.305 ","End":"01:22.320","Text":"N minus P is going to be 2 minus 0 and that is equal to 2,"},{"Start":"01:22.320 ","End":"01:24.620","Text":"so that\u0027s the answer."},{"Start":"01:24.620 ","End":"01:30.545","Text":"But it would be helpful possibly for you to have an illustration of this."},{"Start":"01:30.545 ","End":"01:33.260","Text":"This is the contour Gamma,"},{"Start":"01:33.260 ","End":"01:35.750","Text":"the circle of radius 2."},{"Start":"01:35.750 ","End":"01:37.445","Text":"Actually we start here,"},{"Start":"01:37.445 ","End":"01:44.000","Text":"that\u0027s when Theta equals 0 and we go around once anticlockwise."},{"Start":"01:44.000 ","End":"01:47.360","Text":"The zeros are at 0 and 1,"},{"Start":"01:47.360 ","End":"01:50.890","Text":"which are inside, and these go to the 0 here."},{"Start":"01:50.890 ","End":"01:57.919","Text":"Now the function z squared minus z takes this circle to this shape."},{"Start":"01:57.919 ","End":"02:00.695","Text":"You know what? I have an animation to show you,"},{"Start":"02:00.695 ","End":"02:04.140","Text":"and with that we\u0027ll conclude this clip."}],"ID":23764},{"Watched":false,"Name":"Example 3","Duration":"2m 40s","ChapterTopicVideoID":22926,"CourseChapterTopicPlaylistID":102291,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.070","Text":"Here we have an example of this equality and we\u0027ll show that it"},{"Start":"00:05.070 ","End":"00:10.110","Text":"holds in the particular case where f(z) is e^z,"},{"Start":"00:10.110 ","End":"00:14.340","Text":"the exponential function, and Gamma is the unit circle"},{"Start":"00:14.340 ","End":"00:19.155","Text":"traversed once counterclockwise with this parameterization."},{"Start":"00:19.155 ","End":"00:22.140","Text":"We\u0027ll show this by just evaluating the left-hand side,"},{"Start":"00:22.140 ","End":"00:24.000","Text":"evaluating the right-hand side,"},{"Start":"00:24.000 ","End":"00:26.010","Text":"and seeing that they are equal."},{"Start":"00:26.010 ","End":"00:28.950","Text":"We start with the left-hand side."},{"Start":"00:28.950 ","End":"00:31.350","Text":"By the argument principle,"},{"Start":"00:31.350 ","End":"00:33.090","Text":"this is equal to n minus p;"},{"Start":"00:33.090 ","End":"00:38.010","Text":"number of zeros minus the number of poles inside the unit disk."},{"Start":"00:38.010 ","End":"00:42.500","Text":"The exponential function has no zeros and no poles."},{"Start":"00:42.500 ","End":"00:51.055","Text":"N and p are both zeros and n minus p is 0 and that gives us that the left-hand side is 0."},{"Start":"00:51.055 ","End":"00:53.730","Text":"On to the right-hand side,"},{"Start":"00:53.730 ","End":"00:57.495","Text":"Gamma is equal to this."},{"Start":"00:57.495 ","End":"01:00.870","Text":"If z is on this contour,"},{"Start":"01:00.870 ","End":"01:07.880","Text":"then f(z) is e to the power of e to the i Theta."},{"Start":"01:07.880 ","End":"01:14.130","Text":"We can break up e to the i Theta as cosine Theta plus i sine Theta,"},{"Start":"01:14.130 ","End":"01:19.875","Text":"Euler\u0027s formula, and using rules of exponents."},{"Start":"01:19.875 ","End":"01:27.040","Text":"The argument is just the argument of this part because this part is real."},{"Start":"01:27.040 ","End":"01:31.150","Text":"The argument of e to the power of i Alpha, that\u0027s cool."},{"Start":"01:31.150 ","End":"01:32.800","Text":"That is just Alpha."},{"Start":"01:32.800 ","End":"01:38.085","Text":"But strictly speaking, it\u0027s only defined up to 2n Pi."},{"Start":"01:38.085 ","End":"01:47.320","Text":"The increment, the Delta of the argument is the sine of Theta equals 0,"},{"Start":"01:47.320 ","End":"01:50.755","Text":"subtracted from Theta equals 2Pi."},{"Start":"01:50.755 ","End":"01:52.165","Text":"The 2n Pis cancel."},{"Start":"01:52.165 ","End":"01:55.480","Text":"We usually don\u0027t even bother with this because you just know it cancels."},{"Start":"01:55.480 ","End":"01:58.164","Text":"It\u0027s sine 2Pi minus sine of 0,"},{"Start":"01:58.164 ","End":"02:02.240","Text":"which is 0 minus 0, which is 0."},{"Start":"02:02.240 ","End":"02:06.645","Text":"The right-hand side is equal to 0."},{"Start":"02:06.645 ","End":"02:10.310","Text":"We just saw that the left-hand side is 0,"},{"Start":"02:10.310 ","End":"02:13.040","Text":"and so we\u0027ve proven what we had to prove,"},{"Start":"02:13.040 ","End":"02:15.170","Text":"but I\u0027d like to illustrate it."},{"Start":"02:15.170 ","End":"02:22.730","Text":"This is the winding number of the image of Gamma around the 0."},{"Start":"02:22.730 ","End":"02:24.485","Text":"Here\u0027s the picture."},{"Start":"02:24.485 ","End":"02:27.665","Text":"As we traverse this one time,"},{"Start":"02:27.665 ","End":"02:30.275","Text":"this goes round here,"},{"Start":"02:30.275 ","End":"02:33.815","Text":"but it never wraps around the 0."},{"Start":"02:33.815 ","End":"02:36.200","Text":"The winding number is 0,"},{"Start":"02:36.200 ","End":"02:40.260","Text":"and that\u0027s what we got. We\u0027re done."}],"ID":23765},{"Watched":false,"Name":"Exercise 1","Duration":"9m 41s","ChapterTopicVideoID":23159,"CourseChapterTopicPlaylistID":102291,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.250","Text":"In this exercise, we are asked to use the Argument Principle to compute the number of"},{"Start":"00:05.250 ","End":"00:11.685","Text":"zeros of the polynomial z squared plus 1 in the upper half-plane."},{"Start":"00:11.685 ","End":"00:13.920","Text":"Now, in this particular example,"},{"Start":"00:13.920 ","End":"00:17.640","Text":"we can do it more easily without the Argument Principle."},{"Start":"00:17.640 ","End":"00:22.760","Text":"Z-squared plus 1 has zeros when z squared is minus 1."},{"Start":"00:22.760 ","End":"00:25.640","Text":"In other words, z equals plus or minus i."},{"Start":"00:25.640 ","End":"00:28.180","Text":"They\u0027re both simple zeros."},{"Start":"00:28.180 ","End":"00:32.070","Text":"Obviously, only z=i is in the upper half-plane,"},{"Start":"00:32.070 ","End":"00:34.810","Text":"and minus i is in the lower half-plane."},{"Start":"00:34.810 ","End":"00:38.790","Text":"We have just the 1 and then number of zeros N=1."},{"Start":"00:38.790 ","End":"00:40.835","Text":"I should have written, yeah,"},{"Start":"00:40.835 ","End":"00:43.705","Text":"we want to find N, the number of zeros."},{"Start":"00:43.705 ","End":"00:46.700","Text":"Now, that\u0027s all very well,"},{"Start":"00:46.700 ","End":"00:52.010","Text":"but we want a technique that will work in a lot more cases,"},{"Start":"00:52.010 ","End":"00:55.130","Text":"not just the easy-to-solve polynomials."},{"Start":"00:55.130 ","End":"00:58.175","Text":"We want to use the Argument Principle for this."},{"Start":"00:58.175 ","End":"00:59.945","Text":"We\u0027re going to assume,"},{"Start":"00:59.945 ","End":"01:03.800","Text":"we\u0027re working in general that p doesn\u0027t have any zeros on"},{"Start":"01:03.800 ","End":"01:09.200","Text":"the real line because we\u0027re going to use a semicircular contour."},{"Start":"01:09.200 ","End":"01:11.990","Text":"There\u0027s the picture, if we\u0027re going to do a contour like this."},{"Start":"01:11.990 ","End":"01:14.480","Text":"We don\u0027t want any zeros on the reels."},{"Start":"01:14.480 ","End":"01:17.659","Text":"We consider our usual semicircular integral,"},{"Start":"01:17.659 ","End":"01:22.520","Text":"which is a semicircle and a straight line segment."},{"Start":"01:22.520 ","End":"01:28.834","Text":"The parameterization of the semicircle part is this."},{"Start":"01:28.834 ","End":"01:32.150","Text":"This line segment part,"},{"Start":"01:32.150 ","End":"01:35.670","Text":"Gamma R is like so."},{"Start":"01:35.670 ","End":"01:38.430","Text":"Eventually we\u0027re going to let R go to infinity."},{"Start":"01:38.430 ","End":"01:42.590","Text":"Let\u0027s assume at the start that R is big enough that all the zeros in"},{"Start":"01:42.590 ","End":"01:47.545","Text":"the upper half-plane are inside this semicircle."},{"Start":"01:47.545 ","End":"01:51.035","Text":"Now, of course, the polynomial has no poles,"},{"Start":"01:51.035 ","End":"01:56.630","Text":"so that p=0 in any domain,"},{"Start":"01:56.630 ","End":"02:01.385","Text":"in particular inside the semicircle of radius R, whatever R is."},{"Start":"02:01.385 ","End":"02:04.775","Text":"We have no zeros or poles on the boundary."},{"Start":"02:04.775 ","End":"02:13.220","Text":"Then we can use the Argument Principle and say that this integral is equal to N minus P,"},{"Start":"02:13.220 ","End":"02:16.210","Text":"but that\u0027s just N because P=0."},{"Start":"02:16.210 ","End":"02:19.610","Text":"That\u0027s also equal to the Delta argument,"},{"Start":"02:19.610 ","End":"02:26.060","Text":"like in the tutorial on the geometric meaning of the Argument Principle."},{"Start":"02:26.060 ","End":"02:28.115","Text":"First, just work on this bit."},{"Start":"02:28.115 ","End":"02:34.130","Text":"This gives us N. We need to compute the increase in the argument"},{"Start":"02:34.130 ","End":"02:41.360","Text":"of P as z traverses along Gamma R and then divide that by 2Pi."},{"Start":"02:41.740 ","End":"02:48.680","Text":"This is equal to break up the Delta into the increase of the argument"},{"Start":"02:48.680 ","End":"02:51.620","Text":"along C_R plus the increase along"},{"Start":"02:51.620 ","End":"02:55.560","Text":"Gamma R. The total increase is the some of the increases."},{"Start":"02:55.560 ","End":"03:01.730","Text":"We have this and then letting R go to infinity, we get this."},{"Start":"03:01.730 ","End":"03:03.935","Text":"Now, next we\u0027ll show 2 things."},{"Start":"03:03.935 ","End":"03:08.310","Text":"First of all, that the limit as R goes to"},{"Start":"03:08.310 ","End":"03:14.420","Text":"infinity of Delta R along the semicircular part is 2Pi."},{"Start":"03:14.420 ","End":"03:17.960","Text":"That\u0027s just for our particular polynomial."},{"Start":"03:17.960 ","End":"03:23.570","Text":"But more generally, the limit as R goes to infinity of the increase of"},{"Start":"03:23.570 ","End":"03:30.375","Text":"the argument of P(z) as z goes along Gamma R is 0."},{"Start":"03:30.375 ","End":"03:33.810","Text":"Once we have these 2 things proven,"},{"Start":"03:33.810 ","End":"03:38.795","Text":"then we can say that N is equal to this plus this,"},{"Start":"03:38.795 ","End":"03:41.855","Text":"and this bit we\u0027ll show is 2Pi,"},{"Start":"03:41.855 ","End":"03:43.460","Text":"this bit is 0."},{"Start":"03:43.460 ","End":"03:46.670","Text":"Altogether, we\u0027ll get the answer of 1,"},{"Start":"03:46.670 ","End":"03:49.190","Text":"which is what we expected because we counted it."},{"Start":"03:49.190 ","End":"03:52.895","Text":"There\u0027s only that 1, 0, and plus i."},{"Start":"03:52.895 ","End":"03:56.335","Text":"It remains to show 1 and 2."},{"Start":"03:56.335 ","End":"04:03.350","Text":"Let\u0027s first show that this limit is 2Pi then we\u0027ll do the other one with Gamma R. If z"},{"Start":"04:03.350 ","End":"04:10.459","Text":"is on this semicircle C_R then z is R_e to the i Theta the exponential form,"},{"Start":"04:10.459 ","End":"04:12.075","Text":"like a polar form."},{"Start":"04:12.075 ","End":"04:16.550","Text":"Theta is between 0 and Pi because it\u0027s only a half circle."},{"Start":"04:16.550 ","End":"04:19.640","Text":"Now, P(z) is z squared plus 1."},{"Start":"04:19.640 ","End":"04:21.200","Text":"If we plug that in,"},{"Start":"04:21.200 ","End":"04:26.835","Text":"we get P(Re) to the i Theta equals R^2 e to the i 2 Theta plus 1."},{"Start":"04:26.835 ","End":"04:31.635","Text":"Then we can take the R^2 e to the i2 Theta outside the brackets."},{"Start":"04:31.635 ","End":"04:35.520","Text":"This is what we get except that the"},{"Start":"04:35.520 ","End":"04:41.525","Text":"exponential moved it from the denominator to the numerator and put a minus there."},{"Start":"04:41.525 ","End":"04:43.280","Text":"Now if you look at this,"},{"Start":"04:43.280 ","End":"04:52.759","Text":"this goes to 0 as R goes to infinity because e to the i something is on the unit circle."},{"Start":"04:52.759 ","End":"04:57.475","Text":"Something bounded times something that goes to 0 goes to 0."},{"Start":"04:57.475 ","End":"05:00.195","Text":"This is true for all Theta."},{"Start":"05:00.195 ","End":"05:03.215","Text":"If we just give this a name, this part here,"},{"Start":"05:03.215 ","End":"05:06.395","Text":"call it little h(R) and Theta."},{"Start":"05:06.395 ","End":"05:07.765","Text":"Then h of R,"},{"Start":"05:07.765 ","End":"05:12.140","Text":"Theta goes to 1 as R goes to infinity."},{"Start":"05:12.140 ","End":"05:14.930","Text":"This is for all Theta and 1,"},{"Start":"05:14.930 ","End":"05:20.645","Text":"we can write in exponential form as 1e to the i times 0."},{"Start":"05:20.645 ","End":"05:24.070","Text":"That\u0027s the argument and that\u0027s the modulus."},{"Start":"05:24.070 ","End":"05:28.745","Text":"We can write h as exponential form."},{"Start":"05:28.745 ","End":"05:34.310","Text":"The modulus would be some little r depending on r and Theta."},{"Start":"05:34.310 ","End":"05:38.885","Text":"The argument, we call that Epsilon of R and Theta."},{"Start":"05:38.885 ","End":"05:42.140","Text":"We can choose it so that these 2 are continuous."},{"Start":"05:42.140 ","End":"05:45.140","Text":"What we can deduce from this limit is that"},{"Start":"05:45.140 ","End":"05:49.220","Text":"the modulus goes to the modulus and the argument goes to the argument."},{"Start":"05:49.220 ","End":"05:52.224","Text":"So for all Theta between 0 and Pi,"},{"Start":"05:52.224 ","End":"05:55.290","Text":"the limit of r R and Theta is 1."},{"Start":"05:55.290 ","End":"05:59.809","Text":"That\u0027s the 1 here. The limit of the Epsilon is 0."},{"Start":"05:59.809 ","End":"06:04.430","Text":"Strictly speaking, it could be 0 plus a multiple of 2Pi,"},{"Start":"06:04.430 ","End":"06:09.980","Text":"but we could always adjust it if it\u0027s continuous and discrete has to be constant,"},{"Start":"06:09.980 ","End":"06:13.325","Text":"has to be the same 2NPi and we can just drop it."},{"Start":"06:13.325 ","End":"06:14.795","Text":"We have these 2 limits."},{"Start":"06:14.795 ","End":"06:19.490","Text":"Now, recall that z itself is R_e to the i Theta."},{"Start":"06:19.490 ","End":"06:20.630","Text":"Where did we see that?"},{"Start":"06:20.630 ","End":"06:24.145","Text":"Here. For z on the semicircle."},{"Start":"06:24.145 ","End":"06:26.390","Text":"We showed that for this P(z),"},{"Start":"06:26.390 ","End":"06:28.860","Text":"here it is."},{"Start":"06:28.860 ","End":"06:30.780","Text":"Just copied that."},{"Start":"06:30.780 ","End":"06:38.120","Text":"If we take h from here and we combine the modulus and the argument,"},{"Start":"06:38.120 ","End":"06:42.550","Text":"we get R^2 from here,"},{"Start":"06:42.550 ","End":"06:44.790","Text":"r of R and Theta from here."},{"Start":"06:44.790 ","End":"06:50.100","Text":"Then we have e to the i2 Theta and e to the i Epsilon,"},{"Start":"06:50.100 ","End":"06:53.135","Text":"e to the i 2 Theta plus Epsilon."},{"Start":"06:53.135 ","End":"06:58.460","Text":"The argument of P(z) is just this exponent here,"},{"Start":"06:58.460 ","End":"07:01.285","Text":"just the part that\u0027s in brackets."},{"Start":"07:01.285 ","End":"07:07.100","Text":"The limit as R goes to infinity of Delta arg of"},{"Start":"07:07.100 ","End":"07:13.005","Text":"P(z) as z goes across C_R is the limit."},{"Start":"07:13.005 ","End":"07:18.138","Text":"Theta goes from 0 to Pi and of the exponent here,"},{"Start":"07:18.138 ","End":"07:20.935","Text":"2 Theta plus Epsilon of r Theta."},{"Start":"07:20.935 ","End":"07:23.135","Text":"To figure out this Delta,"},{"Start":"07:23.135 ","End":"07:27.620","Text":"we just have to figure out the value of the argument at"},{"Start":"07:27.620 ","End":"07:31.810","Text":"the end and the beginning and subtract,"},{"Start":"07:31.810 ","End":"07:36.275","Text":"plug in 0, plug in Pi and subtract,"},{"Start":"07:36.275 ","End":"07:38.500","Text":"plug in Theta=Pi,"},{"Start":"07:38.500 ","End":"07:42.335","Text":"we get 2Pi plus Epsilon of our Pi."},{"Start":"07:42.335 ","End":"07:48.535","Text":"Plugging 0, we have twice 0 plus Epsilon of R, 0."},{"Start":"07:48.535 ","End":"07:53.185","Text":"What we get is 2Pi minus twice 0 is 2Pi."},{"Start":"07:53.185 ","End":"07:56.660","Text":"The limit we can apply to each of these."},{"Start":"07:56.660 ","End":"07:58.205","Text":"If you look back here,"},{"Start":"07:58.205 ","End":"08:01.565","Text":"we see that for all Theta between 0 and Pi,"},{"Start":"08:01.565 ","End":"08:07.505","Text":"the limit as R goes to infinity of Epsilon of R Theta is 0."},{"Start":"08:07.505 ","End":"08:10.549","Text":"Each of these will be 0."},{"Start":"08:10.549 ","End":"08:14.795","Text":"We\u0027ve got 2Pi plus 0 minus 0, and this is 2Pi."},{"Start":"08:14.795 ","End":"08:16.895","Text":"That\u0027s what we have to show."},{"Start":"08:16.895 ","End":"08:20.005","Text":"That\u0027s the first limit of 2."},{"Start":"08:20.005 ","End":"08:25.010","Text":"Now let\u0027s get on to the other one where we have to show that the limit, or as I said,"},{"Start":"08:25.010 ","End":"08:28.160","Text":"it actually is true without the limit but it doesn\u0027t"},{"Start":"08:28.160 ","End":"08:31.545","Text":"hurt to throw in limit of the increment,"},{"Start":"08:31.545 ","End":"08:37.085","Text":"the increase in the argument of P(z) along Gamma R the straight line part."},{"Start":"08:37.085 ","End":"08:40.145","Text":"That\u0027s 0. That\u0027s what we\u0027re going to show."},{"Start":"08:40.145 ","End":"08:45.785","Text":"Z is on this straight line path that\u0027s on the real axis."},{"Start":"08:45.785 ","End":"08:53.375","Text":"Then z is some x between minus R and r. P(z) is z^2 plus 1."},{"Start":"08:53.375 ","End":"08:55.685","Text":"P(x) is x^2 plus 1,"},{"Start":"08:55.685 ","End":"08:57.395","Text":"and x is real."},{"Start":"08:57.395 ","End":"09:01.430","Text":"This is bigger than 0, strictly positive."},{"Start":"09:01.430 ","End":"09:07.490","Text":"That means that the argument of P(z) for z on"},{"Start":"09:07.490 ","End":"09:14.485","Text":"Gamma R is 0 because the argument of a positive real number is 0."},{"Start":"09:14.485 ","End":"09:19.220","Text":"The increment along the Gamma R is 0."},{"Start":"09:19.220 ","End":"09:21.650","Text":"If something\u0027s constant, it doesn\u0027t change,"},{"Start":"09:21.650 ","End":"09:23.405","Text":"and so it\u0027s 0."},{"Start":"09:23.405 ","End":"09:28.570","Text":"If we just throw in limit as R goes to infinity, it\u0027s still 0."},{"Start":"09:28.570 ","End":"09:31.580","Text":"That\u0027s the second of the 2 limits we had to show."},{"Start":"09:31.580 ","End":"09:34.140","Text":"This was easier than the first one."},{"Start":"09:36.430 ","End":"09:40.830","Text":"That completes the exercise. We\u0027re done."}],"ID":23988},{"Watched":false,"Name":"Exercise 2","Duration":"9m 30s","ChapterTopicVideoID":23160,"CourseChapterTopicPlaylistID":102291,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.230","Text":"In this exercise, we\u0027ll use the Argument Principle to compute the number of"},{"Start":"00:04.230 ","End":"00:10.440","Text":"zeros of this polynomial that are in the right-half plane."},{"Start":"00:10.440 ","End":"00:13.920","Text":"What we\u0027ll do, because it\u0027s right-half plane,"},{"Start":"00:13.920 ","End":"00:17.100","Text":"is take a right semicircular contour."},{"Start":"00:17.100 ","End":"00:18.779","Text":"See the picture first."},{"Start":"00:18.779 ","End":"00:21.195","Text":"What we\u0027ll do is for large R,"},{"Start":"00:21.195 ","End":"00:23.190","Text":"we\u0027ll consider the semicircular contour,"},{"Start":"00:23.190 ","End":"00:25.335","Text":"which is made up of 2 pieces,"},{"Start":"00:25.335 ","End":"00:28.500","Text":"a straight line piece and a semicircular piece."},{"Start":"00:28.500 ","End":"00:30.885","Text":"Together that\u0027s big Gamma."},{"Start":"00:30.885 ","End":"00:33.435","Text":"It\u0027s little Gamma plus big C,"},{"Start":"00:33.435 ","End":"00:38.940","Text":"C is the semicircle part and Gamma is the straight line part."},{"Start":"00:38.940 ","End":"00:42.105","Text":"Perhaps I should have used the letter y instead of x, so it doesn\u0027t matter."},{"Start":"00:42.105 ","End":"00:49.415","Text":"Later, we\u0027ll show that p(z) doesn\u0027t have any zeros on the imaginary line itself."},{"Start":"00:49.415 ","End":"00:55.740","Text":"No zeros on Gamma R. I will choose R big enough so that"},{"Start":"00:55.740 ","End":"01:03.994","Text":"all the zeros of p(z) that are in the right-half plane are inside the contour big Gamma."},{"Start":"01:03.994 ","End":"01:05.570","Text":"There\u0027s only a finite number of them."},{"Start":"01:05.570 ","End":"01:07.750","Text":"In fact, we know there\u0027s exactly 4 zeros,"},{"Start":"01:07.750 ","End":"01:13.550","Text":"so we find that number so we can get R big enough to encompass all the ones on the right."},{"Start":"01:13.550 ","End":"01:15.079","Text":"Note that P is a polynomial,"},{"Start":"01:15.079 ","End":"01:16.430","Text":"so it has no poles at all."},{"Start":"01:16.430 ","End":"01:23.220","Text":"In particular, it has no poles in the right-half plane and no poles on C_R itself."},{"Start":"01:23.870 ","End":"01:27.920","Text":"We can use the Argument Principle and get"},{"Start":"01:27.920 ","End":"01:33.020","Text":"that this integral is number of zeros minus number of poles,"},{"Start":"01:33.020 ","End":"01:34.790","Text":"but poles equals 0,"},{"Start":"01:34.790 ","End":"01:36.740","Text":"so it\u0027s just number of zeros."},{"Start":"01:36.740 ","End":"01:43.385","Text":"We know that this is also equal to the increase in the argument divided by 2Pi."},{"Start":"01:43.385 ","End":"01:46.385","Text":"We have a formula for N, which is this,"},{"Start":"01:46.385 ","End":"01:50.780","Text":"and we can break up this argument difference into 2 bits."},{"Start":"01:50.780 ","End":"01:56.330","Text":"The change in argument on the semicircle and the change in argument on"},{"Start":"01:56.330 ","End":"01:59.365","Text":"the straight line part and C_R and"},{"Start":"01:59.365 ","End":"02:04.250","Text":"Gamma R. If we take the limit as R goes to infinity of both sides,"},{"Start":"02:04.250 ","End":"02:06.140","Text":"this is a constant, doesn\u0027t depend on R,"},{"Start":"02:06.140 ","End":"02:10.760","Text":"so we just put a limit here and here and we have this."},{"Start":"02:10.760 ","End":"02:15.510","Text":"What we\u0027ll do is we\u0027ll show that, 2 things,"},{"Start":"02:15.510 ","End":"02:22.800","Text":"that this limit here is 4Pi and the other limit is 0."},{"Start":"02:22.800 ","End":"02:25.110","Text":"If we show these 2 things,"},{"Start":"02:25.110 ","End":"02:32.525","Text":"then we\u0027ll get that N by substituting in here is 1/2Pi times 4Pi plus 1/2Pi times 0."},{"Start":"02:32.525 ","End":"02:35.285","Text":"It will come out to be exactly 2."},{"Start":"02:35.285 ","End":"02:37.639","Text":"We have to show these 2 things."},{"Start":"02:37.639 ","End":"02:41.725","Text":"This is the first of the 2 limits that we have to show."},{"Start":"02:41.725 ","End":"02:44.410","Text":"This one should come out to 4Pi,"},{"Start":"02:44.410 ","End":"02:46.070","Text":"which is a bit computational."},{"Start":"02:46.070 ","End":"02:48.980","Text":"Let z be on the semicircle C_R."},{"Start":"02:48.980 ","End":"02:50.210","Text":"Then in exponential form,"},{"Start":"02:50.210 ","End":"02:52.475","Text":"z is Re to the R Theta."},{"Start":"02:52.475 ","End":"02:57.350","Text":"The argument is between minus Pi/2 and Pi/2."},{"Start":"02:57.350 ","End":"03:00.695","Text":"The polynomial recall is this."},{"Start":"03:00.695 ","End":"03:05.800","Text":"Substituting Re to the R Theta instead of z, we get this."},{"Start":"03:05.800 ","End":"03:10.875","Text":"Now, let\u0027s take R^4 e^R 4 Theta outside the brackets."},{"Start":"03:10.875 ","End":"03:12.785","Text":"That\u0027s what we\u0027re left with."},{"Start":"03:12.785 ","End":"03:18.505","Text":"Let\u0027s call this bit in brackets a function h of R and Theta."},{"Start":"03:18.505 ","End":"03:24.420","Text":"The limit as R goes to infinity of h R, Theta is 1."},{"Start":"03:24.420 ","End":"03:27.210","Text":"All these go to 0,"},{"Start":"03:27.210 ","End":"03:29.700","Text":"so we\u0027re just left with 1 and in an exponential form that\u0027s"},{"Start":"03:29.700 ","End":"03:36.970","Text":"1e^i 0 could take multiple of 2Pi, we will take 0."},{"Start":"03:37.060 ","End":"03:40.850","Text":"Now, this in exponential form,"},{"Start":"03:40.850 ","End":"03:45.230","Text":"we can write as little re^i Epsilon"},{"Start":"03:45.230 ","End":"03:49.955","Text":"where r and Epsilon are functions of big R and Theta and continuous."},{"Start":"03:49.955 ","End":"03:53.210","Text":"The limit here of this being 1,"},{"Start":"03:53.210 ","End":"03:57.230","Text":"the modulus goes to 1 and the argument goes to 0,"},{"Start":"03:57.230 ","End":"03:59.150","Text":"and that\u0027s what I\u0027ve written here."},{"Start":"03:59.150 ","End":"04:05.865","Text":"Now, Z is Re^i Theta in exponential form for z on the semicircle."},{"Start":"04:05.865 ","End":"04:10.930","Text":"P(z) is equal to this from here."},{"Start":"04:11.120 ","End":"04:18.020","Text":"We can write this as this by combining the modulus times"},{"Start":"04:18.020 ","End":"04:24.164","Text":"the modulus here and the arguments we add plus what we get."},{"Start":"04:24.164 ","End":"04:25.640","Text":"Back to this limit,"},{"Start":"04:25.640 ","End":"04:32.415","Text":"we have that the argument of p(z) is this here."},{"Start":"04:32.415 ","End":"04:40.595","Text":"The Delta of the argument is the Delta when Theta goes from minus Pi/2 to Pi/2."},{"Start":"04:40.595 ","End":"04:43.375","Text":"Then we want the limit as R goes to infinity."},{"Start":"04:43.375 ","End":"04:49.213","Text":"This Delta, we can compute by substituting the upper and lower limits of Theta,"},{"Start":"04:49.213 ","End":"04:53.420","Text":"and that will give us 4 times Pi/2 plus Epsilon of R,"},{"Start":"04:53.420 ","End":"04:57.565","Text":"Pi/2, and so on with minus Pi/2."},{"Start":"04:57.565 ","End":"05:04.545","Text":"This is equal to 4Pi/2 minus minus 4Pi/2 gives us 4Pi."},{"Start":"05:04.545 ","End":"05:06.375","Text":"Then the limits we can put in here."},{"Start":"05:06.375 ","End":"05:10.080","Text":"We have the limit of Epsilon of R Theta,"},{"Start":"05:10.080 ","End":"05:13.775","Text":"one\u0027s with Pi/2 and one\u0027s with minus Pi/2 subtracted."},{"Start":"05:13.775 ","End":"05:16.385","Text":"However, for all Theta,"},{"Start":"05:16.385 ","End":"05:20.905","Text":"this limit of Epsilon R Theta is 0."},{"Start":"05:20.905 ","End":"05:23.160","Text":"What did we say that?"},{"Start":"05:23.160 ","End":"05:28.335","Text":"Here. We get 4Pi plus 0 minus 0, which is 4Pi."},{"Start":"05:28.335 ","End":"05:30.410","Text":"That\u0027s the first limit we had to show."},{"Start":"05:30.410 ","End":"05:32.230","Text":"Now, on to the second."},{"Start":"05:32.230 ","End":"05:39.590","Text":"This is the second limit that we have to show that the increments in the argument of"},{"Start":"05:39.590 ","End":"05:47.700","Text":"p(z) goes along Gamma R is 0 to the limit as R goes to infinity."},{"Start":"05:47.700 ","End":"05:50.585","Text":"There\u0027s a picture I brought we had earlier,"},{"Start":"05:50.585 ","End":"05:54.315","Text":"we\u0027re going to do this part now, Gamma R,"},{"Start":"05:54.315 ","End":"06:00.260","Text":"so z that\u0027s on this vertical segment is of the form iy,"},{"Start":"06:00.260 ","End":"06:07.140","Text":"where y is between minus R and R. Now p(z) is p(iy),"},{"Start":"06:07.140 ","End":"06:11.540","Text":"and plugging that in to the formula for p(z),"},{"Start":"06:11.540 ","End":"06:13.385","Text":"this is what we get."},{"Start":"06:13.385 ","End":"06:15.530","Text":"If we simplify it,"},{"Start":"06:15.530 ","End":"06:20.525","Text":"collect all the real separately and the imaginary separately,"},{"Start":"06:20.525 ","End":"06:23.915","Text":"imaginary comes from here and from here."},{"Start":"06:23.915 ","End":"06:26.210","Text":"Here we have i^3,"},{"Start":"06:26.210 ","End":"06:27.440","Text":"which is minus i,"},{"Start":"06:27.440 ","End":"06:29.290","Text":"so that\u0027s this here."},{"Start":"06:29.290 ","End":"06:31.850","Text":"Here we just get the 8iy,"},{"Start":"06:31.850 ","End":"06:34.820","Text":"so real and the imaginary parts."},{"Start":"06:34.820 ","End":"06:38.030","Text":"We can complete the square on this bit and write it"},{"Start":"06:38.030 ","End":"06:41.980","Text":"as follows as something squared plus 3/4."},{"Start":"06:41.980 ","End":"06:45.560","Text":"The real part of p(z) is this bit here,"},{"Start":"06:45.560 ","End":"06:47.600","Text":"and that\u0027s strictly positive."},{"Start":"06:47.600 ","End":"06:49.790","Text":"It\u0027s bigger or equal to 3/4."},{"Start":"06:49.790 ","End":"06:54.800","Text":"The imaginary part is 8y minus 8y^3,"},{"Start":"06:54.800 ","End":"06:57.650","Text":"because the real part is positive,"},{"Start":"06:57.650 ","End":"07:03.230","Text":"p of the curve Gamma R is completely in the right half plane."},{"Start":"07:03.230 ","End":"07:07.460","Text":"We could say already that it can\u0027t wrap around the 0."},{"Start":"07:07.460 ","End":"07:11.820","Text":"The increase in the argument is that most Pi has"},{"Start":"07:11.820 ","End":"07:16.355","Text":"to be actually less than Pi because it\u0027s completely in the right half-plane."},{"Start":"07:16.355 ","End":"07:21.155","Text":"Anyway, let\u0027s do it more formally that we could take that shortcut."},{"Start":"07:21.155 ","End":"07:28.295","Text":"We can assume that the argument is between minus Pi/2 and Pi/2."},{"Start":"07:28.295 ","End":"07:31.300","Text":"We have a choice up to multiples of 2 and Pi,"},{"Start":"07:31.300 ","End":"07:35.350","Text":"but we\u0027ll just take it so that it\u0027s between minus Pi/2 and Pi/2,"},{"Start":"07:35.350 ","End":"07:37.925","Text":"the branch of the argument."},{"Start":"07:37.925 ","End":"07:45.420","Text":"The argument is the arc tangent of the y/x of p(z)."},{"Start":"07:45.420 ","End":"07:54.384","Text":"We can get the Delta of the argument just by subtracting the end minus the start."},{"Start":"07:54.384 ","End":"07:57.760","Text":"If we do that using this formula here, you know what?"},{"Start":"07:57.760 ","End":"08:00.950","Text":"I\u0027ll always leave it to check the computations."},{"Start":"08:00.950 ","End":"08:04.000","Text":"If we let R go to infinity,"},{"Start":"08:04.000 ","End":"08:07.965","Text":"then this goes to 0 because we have a Degree 3"},{"Start":"08:07.965 ","End":"08:12.725","Text":"polynomial over a Degree 4 polynomial and arc tangent into zero is 0."},{"Start":"08:12.725 ","End":"08:14.405","Text":"Similarly for this one,"},{"Start":"08:14.405 ","End":"08:17.690","Text":"so we have 0 minus 0, which is 0."},{"Start":"08:17.690 ","End":"08:19.340","Text":"That\u0027s what we had to show."},{"Start":"08:19.340 ","End":"08:21.545","Text":"I\u0027ll leave you with a picture."},{"Start":"08:21.545 ","End":"08:25.405","Text":"In this picture we take R equals 100."},{"Start":"08:25.405 ","End":"08:28.610","Text":"This is the curve green bit and the red bit,"},{"Start":"08:28.610 ","End":"08:30.080","Text":"and the green goes to the green,"},{"Start":"08:30.080 ","End":"08:31.385","Text":"the red to the red."},{"Start":"08:31.385 ","End":"08:34.220","Text":"This actually doesn\u0027t touch the origin because as we said,"},{"Start":"08:34.220 ","End":"08:38.335","Text":"this red part is completely in the right half-plane."},{"Start":"08:38.335 ","End":"08:41.210","Text":"You know what? I\u0027ll end with an animation,"},{"Start":"08:41.210 ","End":"08:44.070","Text":"but basically we\u0027re done."}],"ID":23989}],"Thumbnail":null,"ID":102291},{"Name":"Rouche s Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Rouche\u0027s Theorem (with proof)","Duration":"9m 59s","ChapterTopicVideoID":22911,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.180","Text":"A new topic, Rouche\u0027s theorem."},{"Start":"00:03.180 ","End":"00:06.240","Text":"I Just added a picture of Rouche."},{"Start":"00:06.240 ","End":"00:09.360","Text":"What his theorem says is as follows,"},{"Start":"00:09.360 ","End":"00:14.355","Text":"we have a region D which is enclosed by a simple contour Gamma,"},{"Start":"00:14.355 ","End":"00:18.540","Text":"so Gamma is the boundary of D. Then we have 2 functions,"},{"Start":"00:18.540 ","End":"00:23.355","Text":"f and g. Both of them are holomorphic on D together with its boundary,"},{"Start":"00:23.355 ","End":"00:28.080","Text":"which is the closure of D. Here\u0027s the important condition."},{"Start":"00:28.080 ","End":"00:35.925","Text":"The absolute value of g is strictly less than the absolute value of f on the boundary,"},{"Start":"00:35.925 ","End":"00:37.960","Text":"that is on the contour."},{"Start":"00:37.960 ","End":"00:46.865","Text":"In this case, the 2 functions f and f plus g have the same number of 0s in D,"},{"Start":"00:46.865 ","End":"00:49.385","Text":"just the same number."},{"Start":"00:49.385 ","End":"00:51.875","Text":"I\u0027ll say more about this in a moment."},{"Start":"00:51.875 ","End":"00:57.995","Text":"The important thing is that we count the 0s together with their order or multiplicity."},{"Start":"00:57.995 ","End":"01:02.285","Text":"Now, this doesn\u0027t mean that they have any 0s in common,"},{"Start":"01:02.285 ","End":"01:04.190","Text":"or that it has the same configuration,"},{"Start":"01:04.190 ","End":"01:09.720","Text":"f could have a 0 of order 2 and a simple 0."},{"Start":"01:09.720 ","End":"01:13.910","Text":"Total of 3. f plus g might have 3 simple 0s."},{"Start":"01:13.910 ","End":"01:16.040","Text":"It might have a 0 of order 3."},{"Start":"01:16.040 ","End":"01:17.270","Text":"They could be different,"},{"Start":"01:17.270 ","End":"01:18.800","Text":"they could have some 0s in common."},{"Start":"01:18.800 ","End":"01:24.005","Text":"All we know is that the total number with multiplicity is the same for these 2 functions."},{"Start":"01:24.005 ","End":"01:28.100","Text":"I\u0027ll give you an example of how this theorem would be used."},{"Start":"01:28.100 ","End":"01:34.500","Text":"We have to show that this polynomial z^5 plus 3z^3 plus"},{"Start":"01:34.500 ","End":"01:40.915","Text":"7 has exactly 5 0s in the open disc of radius 2 around 0."},{"Start":"01:40.915 ","End":"01:49.220","Text":"The idea is to find the suitable f and g such that g in absolute value is less than f on"},{"Start":"01:49.220 ","End":"01:53.660","Text":"the boundary and such that the 0s of f plus"},{"Start":"01:53.660 ","End":"01:58.790","Text":"g are easier to compute than the 0s of f. In this case, I\u0027ll just tell you how it\u0027s done."},{"Start":"01:58.790 ","End":"02:06.935","Text":"We break it up naturally as z^5 for f and 3z^3 plus 7 for g. What we have here is"},{"Start":"02:06.935 ","End":"02:15.320","Text":"f plus g. Gamma will take as the circle of radius 2 and D would be the interior of that,"},{"Start":"02:15.320 ","End":"02:17.300","Text":"the open disc of radius 2."},{"Start":"02:17.300 ","End":"02:21.170","Text":"Now, for each z on the circle,"},{"Start":"02:21.170 ","End":"02:23.605","Text":"the absolute value of g,"},{"Start":"02:23.605 ","End":"02:26.585","Text":"you can follow this just straightforward computation,"},{"Start":"02:26.585 ","End":"02:33.353","Text":"is 31 on the circle where absolute value of z is 2."},{"Start":"02:33.353 ","End":"02:34.470","Text":"The other one,"},{"Start":"02:34.470 ","End":"02:36.630","Text":"f is z^5,"},{"Start":"02:36.630 ","End":"02:39.045","Text":"and that comes out to be 32."},{"Start":"02:39.045 ","End":"02:41.809","Text":"31 is less than 32,"},{"Start":"02:41.809 ","End":"02:43.980","Text":"so the condition is met."},{"Start":"02:43.980 ","End":"02:51.610","Text":"Now of course, the 0s of f are much easier to compute than the 0s of f plus g,"},{"Start":"02:51.610 ","End":"02:53.020","Text":"which is the original."},{"Start":"02:53.020 ","End":"02:55.810","Text":"Our polynomial, which is f plus g,"},{"Start":"02:55.810 ","End":"03:00.220","Text":"has the same number of 0s as just f, which is z^5."},{"Start":"03:00.220 ","End":"03:02.215","Text":"We know that this is 5."},{"Start":"03:02.215 ","End":"03:06.550","Text":"It has a 0 of order 5, z equals 0."},{"Start":"03:06.550 ","End":"03:11.275","Text":"We know that this polynomial has 5 0s inside this disc."},{"Start":"03:11.275 ","End":"03:15.470","Text":"As long as you count the 0s together with their multiplicity."},{"Start":"03:15.470 ","End":"03:19.930","Text":"We\u0027ve presented Rouche\u0027s theorem and even given an example of it\u0027s use,"},{"Start":"03:19.930 ","End":"03:21.940","Text":"so now it\u0027s time to prove it."},{"Start":"03:21.940 ","End":"03:24.580","Text":"Here goes, f and g,"},{"Start":"03:24.580 ","End":"03:26.330","Text":"or as in the theorem,"},{"Start":"03:26.330 ","End":"03:28.505","Text":"let\u0027s have a new function,"},{"Start":"03:28.505 ","End":"03:30.990","Text":"f(z) plus tg(z),"},{"Start":"03:30.990 ","End":"03:34.535","Text":"where t is some parameter between 0 and 1."},{"Start":"03:34.535 ","End":"03:38.315","Text":"I\u0027m going to show that this function is non-0 on Gamma."},{"Start":"03:38.315 ","End":"03:41.215","Text":"Take some z on Gamma,"},{"Start":"03:41.215 ","End":"03:44.090","Text":"and then we can evaluate f(z) plus"},{"Start":"03:44.090 ","End":"03:48.200","Text":"tg(z) in absolute value by one of the triangle inequalities."},{"Start":"03:48.200 ","End":"03:50.900","Text":"This is bigger or equal to this."},{"Start":"03:50.900 ","End":"03:54.380","Text":"Then we can take the t out because t is"},{"Start":"03:54.380 ","End":"03:59.045","Text":"non-negative and we can drop the outer absolute value."},{"Start":"03:59.045 ","End":"04:03.440","Text":"Then we can say that t is less than or equal to 1 so minus"},{"Start":"04:03.440 ","End":"04:08.090","Text":"t is bigger or equal to minus 1 and get this,"},{"Start":"04:08.090 ","End":"04:10.720","Text":"the 1 we can just throw out."},{"Start":"04:10.720 ","End":"04:15.770","Text":"Since the absolute value of g is less than the absolute value of f,"},{"Start":"04:15.770 ","End":"04:19.175","Text":"this comes out to be strictly positive."},{"Start":"04:19.175 ","End":"04:25.370","Text":"This f plus tg is also holomorphic because each of f and g is,"},{"Start":"04:25.370 ","End":"04:28.630","Text":"and it\u0027s just the addition and multiplication by a constant."},{"Start":"04:28.630 ","End":"04:35.900","Text":"It\u0027s holomorphic, so it has no poles on the closure of D and in particular on Gamma,"},{"Start":"04:35.900 ","End":"04:39.100","Text":"the contour which is inside D bar."},{"Start":"04:39.100 ","End":"04:45.160","Text":"It also has no 0s because everywhere it\u0027s bigger than 0 on Gamma."},{"Start":"04:45.160 ","End":"04:49.900","Text":"That means that we can use the argument principle."},{"Start":"04:49.900 ","End":"04:57.935","Text":"The number of 0s of f plus tg is given by this formula."},{"Start":"04:57.935 ","End":"05:01.010","Text":"It\u0027s the derivative over the function itself and the"},{"Start":"05:01.010 ","End":"05:04.820","Text":"integral along the contour then divided by 2Pi i."},{"Start":"05:04.820 ","End":"05:08.525","Text":"That\u0027s the number of 0s in the domain D. Now,"},{"Start":"05:08.525 ","End":"05:11.345","Text":"let\u0027s forget that t is a parameter,"},{"Start":"05:11.345 ","End":"05:16.500","Text":"let\u0027s give it a status of a variable and t can vary from 0-1."},{"Start":"05:16.500 ","End":"05:19.190","Text":"We can look at it as a function of 2 variables,"},{"Start":"05:19.190 ","End":"05:23.915","Text":"the real variable t and a complex variable z."},{"Start":"05:23.915 ","End":"05:28.895","Text":"Define it on z belongs to Gamma and t between 0 and 1."},{"Start":"05:28.895 ","End":"05:32.510","Text":"H is continuous, the denominator is not 0"},{"Start":"05:32.510 ","End":"05:37.025","Text":"and the numerator and the denominator are analytic, therefore continuous."},{"Start":"05:37.025 ","End":"05:39.530","Text":"It\u0027s continuous on a compact set."},{"Start":"05:39.530 ","End":"05:42.110","Text":"This is the product of compact sets,"},{"Start":"05:42.110 ","End":"05:46.305","Text":"and therefore H is uniformly continuous."},{"Start":"05:46.305 ","End":"05:49.250","Text":"If it\u0027s uniformly continuous,"},{"Start":"05:49.250 ","End":"05:53.810","Text":"then it\u0027s the theorem that the integral with respect to one of"},{"Start":"05:53.810 ","End":"05:58.916","Text":"the variables is a continuous function in the remaining variable."},{"Start":"05:58.916 ","End":"06:00.815","Text":"Here we take the integral with respect to z,"},{"Start":"06:00.815 ","End":"06:06.790","Text":"so we get a function of t. It\u0027s continuous on 0,1."},{"Start":"06:06.790 ","End":"06:11.680","Text":"Now, H is integer valued because this is"},{"Start":"06:11.680 ","End":"06:18.008","Text":"the number of 0s minus the number of poles of f plus tg."},{"Start":"06:18.008 ","End":"06:21.010","Text":"Now, an integer valued function on"},{"Start":"06:21.010 ","End":"06:26.440","Text":"a connected set is constant and if it\u0027s constant on this interval,"},{"Start":"06:26.440 ","End":"06:29.470","Text":"then that means that I can plug in any 2 values,"},{"Start":"06:29.470 ","End":"06:31.235","Text":"in particular 0 and 1,"},{"Start":"06:31.235 ","End":"06:33.160","Text":"and they are going to be equal."},{"Start":"06:33.160 ","End":"06:35.980","Text":"If we plug in t=0,"},{"Start":"06:35.980 ","End":"06:37.585","Text":"we get this expression."},{"Start":"06:37.585 ","End":"06:39.550","Text":"I guess we could see it up here, t=0,"},{"Start":"06:39.550 ","End":"06:42.545","Text":"we get f\u0027(f) and if t=1,"},{"Start":"06:42.545 ","End":"06:45.205","Text":"we get f plus g,"},{"Start":"06:45.205 ","End":"06:47.440","Text":"the derivative over the function,"},{"Start":"06:47.440 ","End":"06:50.530","Text":"1 over 2Pi i times the integral."},{"Start":"06:50.530 ","End":"06:56.255","Text":"That means that f and f plus g have the same number of 0s."},{"Start":"06:56.255 ","End":"07:00.980","Text":"This integral is the number of 0s minus poles of f,"},{"Start":"07:00.980 ","End":"07:02.570","Text":"but f has no poles,"},{"Start":"07:02.570 ","End":"07:08.150","Text":"it\u0027s analytic, and this is the number of 0s minus poles of f plus g. But again,"},{"Start":"07:08.150 ","End":"07:09.395","Text":"this has no poles,"},{"Start":"07:09.395 ","End":"07:11.239","Text":"so it\u0027s just the number of 0s."},{"Start":"07:11.239 ","End":"07:14.120","Text":"This is what we had to show that these 2 functions,"},{"Start":"07:14.120 ","End":"07:19.775","Text":"f and f plus g have the same number of 0s in D. That concludes the proof."},{"Start":"07:19.775 ","End":"07:25.625","Text":"Next, I\u0027m going to show you an alternative proof and you\u0027re free to watch it or not."},{"Start":"07:25.625 ","End":"07:27.920","Text":"First, let\u0027s get back to the theorem."},{"Start":"07:27.920 ","End":"07:30.980","Text":"Just to remind you of what Rouche\u0027s theorem is,"},{"Start":"07:30.980 ","End":"07:34.105","Text":"just pause this and read it if you want."},{"Start":"07:34.105 ","End":"07:36.725","Text":"Let\u0027s start the alternative proof."},{"Start":"07:36.725 ","End":"07:39.240","Text":"Let h be f plus g,"},{"Start":"07:39.240 ","End":"07:41.685","Text":"where f and g are as in the theorem,"},{"Start":"07:41.685 ","End":"07:45.250","Text":"and then for any z on the contour Gamma,"},{"Start":"07:45.250 ","End":"07:51.935","Text":"we have that absolute value of g is less than absolute value of f, at this point z."},{"Start":"07:51.935 ","End":"07:53.870","Text":"Let\u0027s do some inequalities."},{"Start":"07:53.870 ","End":"07:56.465","Text":"The absolute value of h minus f,"},{"Start":"07:56.465 ","End":"07:58.880","Text":"which is the same as f minus h,"},{"Start":"07:58.880 ","End":"08:04.430","Text":"is less than f because f minus h is g. Next,"},{"Start":"08:04.430 ","End":"08:07.670","Text":"dividing this and this by absolute value of"},{"Start":"08:07.670 ","End":"08:12.545","Text":"f(z) and putting the f(z) inside, we get this."},{"Start":"08:12.545 ","End":"08:17.480","Text":"Then if we let F be h over f,"},{"Start":"08:17.480 ","End":"08:24.560","Text":"then we get that f(z) belongs to the set of all w,"},{"Start":"08:24.560 ","End":"08:30.605","Text":"such that absolute value of w minus 1 is less than 1. w would just be like this."},{"Start":"08:30.605 ","End":"08:36.230","Text":"This is the disc of radius 1 centered at 1."},{"Start":"08:36.230 ","End":"08:41.450","Text":"This means that the image of the contour Gamma under"},{"Start":"08:41.450 ","End":"08:48.375","Text":"the function F is completely inside this open disc centered at 1 with radius 1."},{"Start":"08:48.375 ","End":"08:51.900","Text":"Notice that this disc just misses the 0."},{"Start":"08:51.900 ","End":"08:53.415","Text":"It doesn\u0027t contain 0,"},{"Start":"08:53.415 ","End":"09:00.200","Text":"so the image of the contour Gamma will not wrap around the 0."},{"Start":"09:00.200 ","End":"09:03.820","Text":"The winding number relative to 0 will be 0."},{"Start":"09:03.820 ","End":"09:12.940","Text":"This winding number is 1 over 2Pi Delta arg of F(z) as z traverses the contour Gamma."},{"Start":"09:12.940 ","End":"09:21.715","Text":"But this equals the number of 0s minus the number of poles of F so this difference is 0."},{"Start":"09:21.715 ","End":"09:24.878","Text":"The 0s of F are the 0s of h,"},{"Start":"09:24.878 ","End":"09:33.320","Text":"and the poles of F are the 0s of f. That means that we can replace this."},{"Start":"09:33.320 ","End":"09:39.830","Text":"The 0s of f are the 0s of h the same 0s are the same number of them also."},{"Start":"09:39.830 ","End":"09:43.495","Text":"From here we can replace this with this."},{"Start":"09:43.495 ","End":"09:50.480","Text":"The number of 0s of f plus g is that\u0027s what h is,"},{"Start":"09:50.480 ","End":"09:56.045","Text":"is equal to the number of 0s of f. I just take the minus bring it to the other side."},{"Start":"09:56.045 ","End":"10:00.720","Text":"That proves what we wanted and we\u0027re done."}],"ID":23766},{"Watched":false,"Name":"Exercise 1","Duration":"2m 8s","ChapterTopicVideoID":22912,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.155","Text":"In this exercise we have the function h of z which is z^4 plus 5z plus 1,"},{"Start":"00:07.155 ","End":"00:12.705","Text":"and we want to find the number of 0s it has in the open unit disc."},{"Start":"00:12.705 ","End":"00:17.115","Text":"We\u0027re going to use Rouche\u0027s theorem and I\u0027ll remind you what that is."},{"Start":"00:17.115 ","End":"00:21.000","Text":"Here\u0027s Rouche\u0027s theorem and I\u0027ll let you pause and look at it."},{"Start":"00:21.000 ","End":"00:26.475","Text":"Gamma is the boundary of D which is the unit circle,"},{"Start":"00:26.475 ","End":"00:32.280","Text":"and what we\u0027re going to do is we\u0027re going to break h up into f plus g in"},{"Start":"00:32.280 ","End":"00:39.428","Text":"such a way that the absolute value of g is less than the absolute value of f on Gamma,"},{"Start":"00:39.428 ","End":"00:44.522","Text":"and then by Rouche\u0027s theorem h and f will have the same number of 0s"},{"Start":"00:44.522 ","End":"00:50.780","Text":"and will do it in such a way that the 0s of f are easy to compute."},{"Start":"00:50.780 ","End":"00:56.615","Text":"At least that it\u0027s easy to find the number of 0s because as it is, it\u0027s not easy."},{"Start":"00:56.615 ","End":"01:02.030","Text":"Now the obvious decomposition to take f as z^4 and g is"},{"Start":"01:02.030 ","End":"01:09.170","Text":"5z plus 1 doesn\u0027t work because we can\u0027t get this inequality."},{"Start":"01:09.170 ","End":"01:14.180","Text":"By playing around we come upon the right way to do it which"},{"Start":"01:14.180 ","End":"01:19.155","Text":"is to take f as the 5z part and g is z^4 plus 1."},{"Start":"01:19.155 ","End":"01:22.338","Text":"Certainly the zero of f are unifying,"},{"Start":"01:22.338 ","End":"01:25.790","Text":"let\u0027s just see if we can get that inequality on Gamma."},{"Start":"01:25.790 ","End":"01:28.910","Text":"On Gamma the absolute value of z is 1,"},{"Start":"01:28.910 ","End":"01:34.200","Text":"so the absolute value of f is absolute value of 5z which is 5."},{"Start":"01:34.200 ","End":"01:39.240","Text":"On the other hand the absolute value of g is the absolute value of z^4 plus 1,"},{"Start":"01:39.240 ","End":"01:42.182","Text":"and by the triangle inequality we get this."},{"Start":"01:42.182 ","End":"01:47.870","Text":"Absolute value of z is 1 so we get 2 which is definitely less than 5."},{"Start":"01:47.870 ","End":"01:51.920","Text":"H and f have the same number of 0s in D,"},{"Start":"01:51.920 ","End":"01:54.730","Text":"but f has only a single 0."},{"Start":"01:54.730 ","End":"01:58.670","Text":"That is, z equals 0 and it\u0027s of order 1."},{"Start":"01:58.670 ","End":"02:00.890","Text":"By Rouche\u0027s theorem,"},{"Start":"02:00.890 ","End":"02:05.120","Text":"h also has one 0 in D;"},{"Start":"02:05.120 ","End":"02:08.730","Text":"exactly one, and we are done."}],"ID":23767},{"Watched":false,"Name":"Exercise 2","Duration":"1m 47s","ChapterTopicVideoID":22913,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"In this exercise, we have the function h(z) which is actually a polynomial,"},{"Start":"00:04.800 ","End":"00:06.780","Text":"5z^4 plus z plus 1."},{"Start":"00:06.780 ","End":"00:11.175","Text":"I\u0027m going to find out how many zeros it has in the open unit disc."},{"Start":"00:11.175 ","End":"00:13.260","Text":"We\u0027ll be using Rouche\u0027s theorem,"},{"Start":"00:13.260 ","End":"00:16.080","Text":"which I\u0027ll remind you, well here it is."},{"Start":"00:16.080 ","End":"00:19.275","Text":"You can pause it as long as you need to look at it."},{"Start":"00:19.275 ","End":"00:26.250","Text":"That\u0027s what we need to do is to decompose h into f plus g. Gamma is the boundary of D,"},{"Start":"00:26.250 ","End":"00:27.540","Text":"as in the picture."},{"Start":"00:27.540 ","End":"00:29.910","Text":"We want h equals f plus g,"},{"Start":"00:29.910 ","End":"00:36.000","Text":"such that the absolute value of g is less than the absolute value of f on Gamma."},{"Start":"00:36.000 ","End":"00:39.460","Text":"Then h and f will have the same number of zeros."},{"Start":"00:39.460 ","End":"00:42.230","Text":"We\u0027ve got to decompose it in such a way that it\u0027s easy to find"},{"Start":"00:42.230 ","End":"00:45.555","Text":"the number of zeros of f. By playing around,"},{"Start":"00:45.555 ","End":"00:51.990","Text":"the way to decompose it is to take f as 5z^4 and g(z) to be z plus 1."},{"Start":"00:51.990 ","End":"00:57.020","Text":"Certainly, f plus g is h. Question is does this inequality hold?"},{"Start":"00:57.020 ","End":"00:59.690","Text":"On gamma, the absolute value of z is 1."},{"Start":"00:59.690 ","End":"01:03.180","Text":"What we get is that absolute value of f(z) is"},{"Start":"01:03.180 ","End":"01:05.940","Text":"5 and absolute value of g(z) using"},{"Start":"01:05.940 ","End":"01:09.290","Text":"triangle inequality comes out to be less than or equal to 2,"},{"Start":"01:09.290 ","End":"01:10.835","Text":"which is less than 5."},{"Start":"01:10.835 ","End":"01:13.315","Text":"This is certainly less than this."},{"Start":"01:13.315 ","End":"01:17.890","Text":"That means that f and h have the same number of zeros."},{"Start":"01:17.890 ","End":"01:21.660","Text":"The only zero that f has is the 1 that\u0027s zero,"},{"Start":"01:21.660 ","End":"01:22.800","Text":"but when we count,"},{"Start":"01:22.800 ","End":"01:25.440","Text":"we count it as 4 because it has order 4."},{"Start":"01:25.440 ","End":"01:28.790","Text":"Actually f has 4 zeros in D,"},{"Start":"01:28.790 ","End":"01:31.775","Text":"which means that h also,"},{"Start":"01:31.775 ","End":"01:32.970","Text":"by Rouche\u0027s theorem,"},{"Start":"01:32.970 ","End":"01:37.060","Text":"has 4 zeros in D when we count multiplicity."},{"Start":"01:37.060 ","End":"01:41.510","Text":"Of course, the zeros of h could be all simple zeros"},{"Start":"01:41.510 ","End":"01:45.890","Text":"or a triple 0 and a simple 0 as long as it adds up to 4."},{"Start":"01:45.890 ","End":"01:48.450","Text":"Anyway, we\u0027re done here."}],"ID":23768},{"Watched":false,"Name":"Exercise 3","Duration":"2m 15s","ChapterTopicVideoID":22914,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.850","Text":"In this exercise, we have the function h(z)=z^4 plus 5z plus 4,"},{"Start":"00:05.850 ","End":"00:15.870","Text":"and we want to know how many zeros it has in the disc of radius 2 centered at 0."},{"Start":"00:15.870 ","End":"00:18.180","Text":"When we say number of zeros,"},{"Start":"00:18.180 ","End":"00:20.595","Text":"we\u0027re including the multiplicity."},{"Start":"00:20.595 ","End":"00:24.290","Text":"We going to use Rouche\u0027s is theorem and I\u0027ll remind you,"},{"Start":"00:24.290 ","End":"00:25.970","Text":"here\u0027s the theorem,"},{"Start":"00:25.970 ","End":"00:29.615","Text":"you could pause and review as necessary."},{"Start":"00:29.615 ","End":"00:32.960","Text":"We need to split h up as f plus g,"},{"Start":"00:32.960 ","End":"00:35.075","Text":"as in Rouche\u0027s theorem."},{"Start":"00:35.075 ","End":"00:40.010","Text":"Now, the domain D has a border Gamma,"},{"Start":"00:40.010 ","End":"00:43.385","Text":"which is the circle of radius 2."},{"Start":"00:43.385 ","End":"00:49.280","Text":"We want to split h into f plus g such that the absolute value of g is"},{"Start":"00:49.280 ","End":"00:54.875","Text":"less than the absolute value of f on this boundary contour Gamma."},{"Start":"00:54.875 ","End":"00:57.350","Text":"But we also, for practical reasons,"},{"Start":"00:57.350 ","End":"01:01.910","Text":"want to do it in such a way that the number of zeros for f are easier to"},{"Start":"01:01.910 ","End":"01:06.970","Text":"find because h and f will have the same number of zeros."},{"Start":"01:06.970 ","End":"01:08.505","Text":"By a bit of trial and error,"},{"Start":"01:08.505 ","End":"01:12.795","Text":"the way to do it is to take f(z) is z^4,"},{"Start":"01:12.795 ","End":"01:17.020","Text":"and g(z) is this bit, 5z plus 4."},{"Start":"01:17.020 ","End":"01:20.015","Text":"Let\u0027s check what happens on the boundary."},{"Start":"01:20.015 ","End":"01:23.500","Text":"Absolute value of z is 2."},{"Start":"01:23.510 ","End":"01:29.150","Text":"Absolute value of f(z) on the boundary comes out to be exactly 16,"},{"Start":"01:29.150 ","End":"01:31.310","Text":"whereas absolute value of g(z) using"},{"Start":"01:31.310 ","End":"01:35.323","Text":"the triangle inequality comes out to be less than or equal to 14,"},{"Start":"01:35.323 ","End":"01:38.365","Text":"and so certainly less than 16."},{"Start":"01:38.365 ","End":"01:45.185","Text":"The condition is met and also the zeros of f are easy to find because f(z) is z^4,"},{"Start":"01:45.185 ","End":"01:46.969","Text":"it has a single 0,"},{"Start":"01:46.969 ","End":"01:48.305","Text":"but of order 4."},{"Start":"01:48.305 ","End":"01:54.570","Text":"The number of zeros of f is 4 inside D. By"},{"Start":"01:54.570 ","End":"02:01.475","Text":"the Rouche\u0027s theorem number of zeros of h will also be 4 inside D. There are 4 zeros,"},{"Start":"02:01.475 ","End":"02:06.425","Text":"including multiplicity, could be a couple of zeros of order 2,"},{"Start":"02:06.425 ","End":"02:08.405","Text":"could be a 0 of order 3,"},{"Start":"02:08.405 ","End":"02:10.055","Text":"and 2 simple zeros,"},{"Start":"02:10.055 ","End":"02:15.600","Text":"we don\u0027t know, but there will be 4 in total. That\u0027s it."}],"ID":23769},{"Watched":false,"Name":"Exercise 4","Duration":"4m 1s","ChapterTopicVideoID":22915,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"In this exercise, you want to find the number of 0s of h(z),"},{"Start":"00:04.830 ","End":"00:07.140","Text":"which is actually a polynomial function in z,"},{"Start":"00:07.140 ","End":"00:11.235","Text":"is z to the fifth plus 3z plus 1 in the following domain."},{"Start":"00:11.235 ","End":"00:15.285","Text":"Now if you look at this, you\u0027ll see that it\u0027s actually an annulus."},{"Start":"00:15.285 ","End":"00:17.044","Text":"A picture might help."},{"Start":"00:17.044 ","End":"00:20.615","Text":"It\u0027s the blue part, the annulus."},{"Start":"00:20.615 ","End":"00:24.320","Text":"Now what we\u0027re going do is we\u0027re going to use Rouche\u0027s theorem twice."},{"Start":"00:24.320 ","End":"00:28.240","Text":"We\u0027re going to use it once on the disk of radius 2,"},{"Start":"00:28.240 ","End":"00:30.420","Text":"find the number of 0s in here."},{"Start":"00:30.420 ","End":"00:34.735","Text":"Then we\u0027ll use Rouche\u0027s theorem on the disk of radius 1,"},{"Start":"00:34.735 ","End":"00:37.040","Text":"and we\u0027ll subtract the number of 0s in"},{"Start":"00:37.040 ","End":"00:39.905","Text":"the small disk from the number of 0s in the larger disk,"},{"Start":"00:39.905 ","End":"00:44.900","Text":"assuming that there\u0027s no 0s on the boundary itself,"},{"Start":"00:44.900 ","End":"00:48.275","Text":"which is where absolute value of z is equal to 1."},{"Start":"00:48.275 ","End":"00:50.210","Text":"That\u0027s the strategy."},{"Start":"00:50.210 ","End":"00:57.145","Text":"Let\u0027s start with the number of 0s in the larger disk of radius 2 around 0."},{"Start":"00:57.145 ","End":"01:00.643","Text":"Gamma 2 is as illustrated,"},{"Start":"01:00.643 ","End":"01:03.680","Text":"and we want to break h up into f plus g,"},{"Start":"01:03.680 ","End":"01:07.490","Text":"such that the absolute value of g is less than the absolute value of"},{"Start":"01:07.490 ","End":"01:11.450","Text":"f on the boundary Gamma 2,"},{"Start":"01:11.450 ","End":"01:17.659","Text":"and in such a way that the number of 0s of f is easier to compute."},{"Start":"01:17.659 ","End":"01:21.320","Text":"But a trial and error leads us to the following f and g,"},{"Start":"01:21.320 ","End":"01:25.040","Text":"and let\u0027s check that the condition is met on Gamma 2."},{"Start":"01:25.040 ","End":"01:28.310","Text":"Well, absolute value of z is equal to 2,"},{"Start":"01:28.310 ","End":"01:32.180","Text":"so absolute value of f is 32."},{"Start":"01:32.180 ","End":"01:37.340","Text":"Absolute value of g using triangle inequality is less than or equal to 7,"},{"Start":"01:37.340 ","End":"01:39.830","Text":"which is certainly less than 32."},{"Start":"01:39.830 ","End":"01:42.280","Text":"So the condition is met."},{"Start":"01:42.280 ","End":"01:45.830","Text":"It\u0027s much easier to compute the 0s of f. In fact,"},{"Start":"01:45.830 ","End":"01:49.550","Text":"it\u0027s trivial, it has a single 0 at 0,"},{"Start":"01:49.550 ","End":"01:52.715","Text":"but we have to include the order or multiplicity,"},{"Start":"01:52.715 ","End":"01:57.575","Text":"which is 5 because of the z to the fifth here."},{"Start":"01:57.575 ","End":"02:00.860","Text":"So actually when we count it\u0027s 5."},{"Start":"02:00.860 ","End":"02:07.240","Text":"The number of 0s of h equals the number of 0s of f, which is 5."},{"Start":"02:07.240 ","End":"02:12.660","Text":"Just summarizing. So h has 5 0s in the disc of radius 2."},{"Start":"02:12.660 ","End":"02:17.953","Text":"Next you want to compute the number of 0s in the disk of radius 1,"},{"Start":"02:17.953 ","End":"02:22.195","Text":"and the boundary is where absolute value of z equals 1."},{"Start":"02:22.195 ","End":"02:23.900","Text":"I want to break up h,"},{"Start":"02:23.900 ","End":"02:26.315","Text":"different from the previous breakup."},{"Start":"02:26.315 ","End":"02:30.830","Text":"But again, so that absolute value of g is less than absolute value of f on the boundary,"},{"Start":"02:30.830 ","End":"02:33.200","Text":"only this time the boundary is Gamma 1."},{"Start":"02:33.200 ","End":"02:36.320","Text":"It turns out that what does the trick is f of z equals"},{"Start":"02:36.320 ","End":"02:39.510","Text":"3z and g of z is z to the fifth plus 1."},{"Start":"02:39.510 ","End":"02:42.500","Text":"Let\u0027s check the inequality on Gamma 1."},{"Start":"02:42.500 ","End":"02:44.330","Text":"That\u0027s where absolute value of z is 1."},{"Start":"02:44.330 ","End":"02:48.820","Text":"So the absolute value of f of z comes out to be 3 exactly,"},{"Start":"02:48.820 ","End":"02:54.545","Text":"and the absolute value of g of z using triangle inequality is less than or equal to 2,"},{"Start":"02:54.545 ","End":"02:57.220","Text":"which is less than 3."},{"Start":"02:57.220 ","End":"03:01.635","Text":"F(z) is 3z."},{"Start":"03:01.635 ","End":"03:10.655","Text":"The only 0 of f is z=0 of multiplicity or order 1 in the open unit disc."},{"Start":"03:10.655 ","End":"03:16.595","Text":"By Rouche\u0027s theorem the number of 0s of h in the open unit disc is 1."},{"Start":"03:16.595 ","End":"03:20.645","Text":"So h has 1 0 in the smaller disk."},{"Start":"03:20.645 ","End":"03:23.875","Text":"Now we have to take care of the matter of the boundary."},{"Start":"03:23.875 ","End":"03:25.489","Text":"On Gamma 1,"},{"Start":"03:25.489 ","End":"03:31.340","Text":"absolute value of h is bigger or equal to absolute value of f minus absolute value of g,"},{"Start":"03:31.340 ","End":"03:33.500","Text":"and we already showed that this is less than this,"},{"Start":"03:33.500 ","End":"03:35.615","Text":"so it comes up bigger than 0."},{"Start":"03:35.615 ","End":"03:40.400","Text":"So h can\u0027t be 0 on the border Gamma 1,"},{"Start":"03:40.400 ","End":"03:42.103","Text":"where absolute value of z is 1."},{"Start":"03:42.103 ","End":"03:45.830","Text":"Altogether, when we plug into this formula,"},{"Start":"03:45.830 ","End":"03:48.875","Text":"we found that this is equal to 4."},{"Start":"03:48.875 ","End":"03:51.595","Text":"This is 0 and this is 1."},{"Start":"03:51.595 ","End":"03:55.335","Text":"Just a computation, 5 minus 0 minus 1,"},{"Start":"03:55.335 ","End":"03:57.300","Text":"and it comes out to be 4,"},{"Start":"03:57.300 ","End":"04:01.510","Text":"and that\u0027s the answer. We\u0027re done."}],"ID":23770},{"Watched":false,"Name":"Exercise 5","Duration":"4m 9s","ChapterTopicVideoID":22916,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"In this exercise, we\u0027re going to find the number of zeros of the"},{"Start":"00:04.380 ","End":"00:09.210","Text":"function h(z) equals z to the 4th minus 10z plus 1."},{"Start":"00:09.210 ","End":"00:11.850","Text":"In the domain D,"},{"Start":"00:11.850 ","End":"00:16.200","Text":"which if you think about it is an annulus."},{"Start":"00:16.200 ","End":"00:18.615","Text":"A diagram could help,"},{"Start":"00:18.615 ","End":"00:23.020","Text":"we\u0027re talking about the blue ring shape here."},{"Start":"00:23.510 ","End":"00:27.465","Text":"The plan is to use Rouche\u0027s theorem twice."},{"Start":"00:27.465 ","End":"00:34.560","Text":"First of all, we\u0027ll use it to compute the number of zeros inside the larger open disk."},{"Start":"00:34.560 ","End":"00:40.181","Text":"Then we\u0027ll subtract from it the number of zeros in the smaller open disk,"},{"Start":"00:40.181 ","End":"00:44.105","Text":"and we also have to subtract the number of"},{"Start":"00:44.105 ","End":"00:49.800","Text":"zeros on the contour here which will be 0 as we\u0027ll show."},{"Start":"00:49.800 ","End":"00:54.305","Text":"Basically, I have the difference of 2 Rouche theorem results."},{"Start":"00:54.305 ","End":"00:57.020","Text":"Okay, call this one Gamma 3,"},{"Start":"00:57.020 ","End":"01:00.990","Text":"just to match up with the 3 here which is that circle."},{"Start":"01:00.990 ","End":"01:03.810","Text":"It\u0027s the boundary of this disc."},{"Start":"01:03.810 ","End":"01:09.240","Text":"If you want to split h up into f plus g as in Rouche\u0027s theorem,"},{"Start":"01:09.240 ","End":"01:11.075","Text":"and I\u0027ll remind you what that is."},{"Start":"01:11.075 ","End":"01:15.860","Text":"Here is the theorem which you can pause and read I\u0027ll just remark that"},{"Start":"01:15.860 ","End":"01:20.735","Text":"f plus g here is what we call h in our problem."},{"Start":"01:20.735 ","End":"01:25.555","Text":"There\u0027s some trial and error in to finding the right f and g. In this case,"},{"Start":"01:25.555 ","End":"01:30.105","Text":"we can take f to be z^4th this part,"},{"Start":"01:30.105 ","End":"01:36.420","Text":"and the rest of it will be g. We have to show that on this contour here,"},{"Start":"01:36.420 ","End":"01:39.545","Text":"Gamma 3, we have this inequality."},{"Start":"01:39.545 ","End":"01:41.435","Text":"Let\u0027s evaluate each of them."},{"Start":"01:41.435 ","End":"01:45.640","Text":"Absolute value of f(z) turns out to be exactly 81,"},{"Start":"01:45.640 ","End":"01:48.930","Text":"and absolute value of g(z) we can estimate using"},{"Start":"01:48.930 ","End":"01:52.310","Text":"the triangle inequality and get that"},{"Start":"01:52.310 ","End":"01:56.675","Text":"it\u0027s less than or equal to 31 which is definitely less than 84."},{"Start":"01:56.675 ","End":"01:59.090","Text":"Now, the zeros,"},{"Start":"01:59.090 ","End":"02:03.800","Text":"the zeros (f )are clearly just at z=0 but we have to take into"},{"Start":"02:03.800 ","End":"02:09.365","Text":"account multiplicity or order and that\u0027s 4 because of the z^4th."},{"Start":"02:09.365 ","End":"02:11.760","Text":"Actually, the way we count it,"},{"Start":"02:11.760 ","End":"02:17.030","Text":"there are 4 zeros of h in this disk of radius 3,"},{"Start":"02:17.030 ","End":"02:20.840","Text":"and hence also f has 4 zeros."},{"Start":"02:20.840 ","End":"02:24.295","Text":"Just to summarize, h has 4 zeros in"},{"Start":"02:24.295 ","End":"02:28.680","Text":"the disk of radius 3 counting multiplicity, that\u0027s this part."},{"Start":"02:28.680 ","End":"02:30.990","Text":"Now, we\u0027ll go and do this part,"},{"Start":"02:30.990 ","End":"02:34.380","Text":"the number of zeros in the smaller inner disk."},{"Start":"02:34.380 ","End":"02:37.610","Text":"The boundary is the circle of radius 1,"},{"Start":"02:37.610 ","End":"02:41.660","Text":"and we wanted to break up h=f plus g again,"},{"Start":"02:41.660 ","End":"02:46.670","Text":"the breakup that was before won\u0027t do so we\u0027ll do it differently."},{"Start":"02:46.670 ","End":"02:47.900","Text":"Again, a bit of guesswork,"},{"Start":"02:47.900 ","End":"02:54.515","Text":"but we find that taking f to be minus 10z and g(z) is equal to 4th plus 1 will work."},{"Start":"02:54.515 ","End":"02:56.540","Text":"Let\u0027s see on Gamma 1,"},{"Start":"02:56.540 ","End":"02:58.130","Text":"absolute value of z is 1,"},{"Start":"02:58.130 ","End":"03:02.145","Text":"so the absolute value of f comes out to be 10 everywhere."},{"Start":"03:02.145 ","End":"03:06.200","Text":"The absolute value of g(z) by the triangle inequality"},{"Start":"03:06.200 ","End":"03:10.235","Text":"comes out to be less than or equal to 2 which is definitely less than 10."},{"Start":"03:10.235 ","End":"03:13.850","Text":"We see that the only 0(f )in the disc,"},{"Start":"03:13.850 ","End":"03:20.150","Text":"f is minus 10z is just z=0 but that\u0027s a simple 0,"},{"Start":"03:20.150 ","End":"03:23.595","Text":"in other words, order 1 or multiplicity 1."},{"Start":"03:23.595 ","End":"03:25.260","Text":"By the Rouche theorem,"},{"Start":"03:25.260 ","End":"03:29.330","Text":"the number of zeros(h) is the same as the number of zeros(f) in"},{"Start":"03:29.330 ","End":"03:35.970","Text":"this disk which means that h has 1 zero in this disk."},{"Start":"03:35.970 ","End":"03:37.770","Text":"On this Gamma 1,"},{"Start":"03:37.770 ","End":"03:42.290","Text":"the absolute value of h(z) is the absolute value of f plus g"},{"Start":"03:42.290 ","End":"03:44.600","Text":"and by the triangle inequality is bigger or"},{"Start":"03:44.600 ","End":"03:47.560","Text":"equal to absolute value of f minus absolute value of g,"},{"Start":"03:47.560 ","End":"03:51.485","Text":"and that\u0027s a positive because we already showed that this is less than this."},{"Start":"03:51.485 ","End":"03:54.365","Text":"There aren\u0027t any zeros on that boundary."},{"Start":"03:54.365 ","End":"03:56.120","Text":"Now we can do the computation."},{"Start":"03:56.120 ","End":"04:00.170","Text":"We had this arithmetic to do number of zeros here that turned out to be 4."},{"Start":"04:00.170 ","End":"04:02.330","Text":"This was 0 and this was 1,"},{"Start":"04:02.330 ","End":"04:06.515","Text":"so 4 minus 0 minus 1 comes out to be 3."},{"Start":"04:06.515 ","End":"04:09.780","Text":"That\u0027s our answer, and we\u0027re done."}],"ID":23771},{"Watched":false,"Name":"Exercise 6","Duration":"1m 53s","ChapterTopicVideoID":22917,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.115","Text":"In this exercise, we\u0027re asked to find the number of zeros of the function h(z),"},{"Start":"00:05.115 ","End":"00:07.575","Text":"which is as written here,"},{"Start":"00:07.575 ","End":"00:09.960","Text":"inside the domain D,"},{"Start":"00:09.960 ","End":"00:11.985","Text":"which is the open unit disc,"},{"Start":"00:11.985 ","End":"00:17.115","Text":"also written as D(0,1), it\u0027s illustrated here."},{"Start":"00:17.115 ","End":"00:21.720","Text":"We\u0027re going to be using Rouche\u0027s theorem and I\u0027ll remind you,"},{"Start":"00:21.720 ","End":"00:24.840","Text":"you can just pause and read this and let\u0027s get back to where we"},{"Start":"00:24.840 ","End":"00:29.010","Text":"were continuing and the idea is to use Rouche."},{"Start":"00:29.010 ","End":"00:31.620","Text":"Gamma, the boundary,"},{"Start":"00:31.620 ","End":"00:37.650","Text":"is the unit circle and we want to split h up into f plus g in such a way"},{"Start":"00:37.650 ","End":"00:43.860","Text":"that the absolute value of g is less than absolute value of f on the boundary Gamma,"},{"Start":"00:43.860 ","End":"00:48.980","Text":"and we also want to do it in such a way that the number of zeros of f is"},{"Start":"00:48.980 ","End":"00:54.500","Text":"easy to find because Rouche says that h and f have the same number of zeros."},{"Start":"00:54.500 ","End":"00:56.454","Text":"By trial and error, what works,"},{"Start":"00:56.454 ","End":"01:00.025","Text":"we will take f(z) to be 7z,"},{"Start":"01:00.025 ","End":"01:03.750","Text":"and g(z) will be the rest of it. It\u0027s this."},{"Start":"01:03.750 ","End":"01:07.595","Text":"Let\u0027s see if we can get this inequality on Gamma."},{"Start":"01:07.595 ","End":"01:09.740","Text":"Absolute value of z is 1,"},{"Start":"01:09.740 ","End":"01:14.970","Text":"so the absolute value of f(z) comes out to be 7,"},{"Start":"01:14.970 ","End":"01:17.890","Text":"whereas the absolute value of g(z),"},{"Start":"01:17.890 ","End":"01:22.490","Text":"we can estimate it with a triangle inequality comes out to be less than or equal to"},{"Start":"01:22.490 ","End":"01:28.325","Text":"5.5 and therefore definitely less than 7 so this inequality holds."},{"Start":"01:28.325 ","End":"01:32.720","Text":"Also, zeros of f are easy to find because the only 0 of"},{"Start":"01:32.720 ","End":"01:37.745","Text":"f is 0 and it has order 1 to simple 0."},{"Start":"01:37.745 ","End":"01:40.430","Text":"The number of zeros of f inside D,"},{"Start":"01:40.430 ","End":"01:42.605","Text":"we should say, is 1."},{"Start":"01:42.605 ","End":"01:47.730","Text":"Using Rouche\u0027s theorem, number of zeros for h and f are the same so h"},{"Start":"01:47.730 ","End":"01:53.770","Text":"also has one 0 inside of D and we\u0027re done."}],"ID":23772},{"Watched":false,"Name":"Exercise 7","Duration":"5m 28s","ChapterTopicVideoID":22918,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.760","Text":"In this exercise, we\u0027re going to find the number of"},{"Start":"00:02.760 ","End":"00:07.095","Text":"zeros of the function h(z) which is this,"},{"Start":"00:07.095 ","End":"00:10.695","Text":"in the domain D. If you look at this,"},{"Start":"00:10.695 ","End":"00:13.800","Text":"you\u0027ll see that it\u0027s an annulus effect of the picture."},{"Start":"00:13.800 ","End":"00:22.440","Text":"It\u0027s an annulus that\u0027s the area between the radius 1 and radius 3 circles,"},{"Start":"00:22.440 ","End":"00:25.480","Text":"that\u0027s the blue shaded."},{"Start":"00:25.670 ","End":"00:31.490","Text":"The plan is to count the number of zeros altogether in the disc of"},{"Start":"00:31.490 ","End":"00:37.640","Text":"radius 3 and then subtract the number of zeros in the inner disc of radius 1,"},{"Start":"00:37.640 ","End":"00:41.975","Text":"provided that there are no zeros exactly on the circle."},{"Start":"00:41.975 ","End":"00:44.755","Text":"This is the plan written here."},{"Start":"00:44.755 ","End":"00:48.770","Text":"We\u0027ll show that there are no zeros on this disc."},{"Start":"00:48.770 ","End":"00:51.740","Text":"We just subtract this disc minus this disc,"},{"Start":"00:51.740 ","End":"00:53.605","Text":"the number of zeros they\u0027re in."},{"Start":"00:53.605 ","End":"00:58.935","Text":"Gamma_3 is the boundary of the disc of radius 3."},{"Start":"00:58.935 ","End":"01:06.060","Text":"We\u0027ll be using Rouché\u0027s theorem and we want to break h up into f plus g. You know what,"},{"Start":"01:06.060 ","End":"01:08.025","Text":"I\u0027ll remind you of Rouché\u0027s theorem."},{"Start":"01:08.025 ","End":"01:13.310","Text":"Here it is, I\u0027ll leave you to pause the video and study this."},{"Start":"01:13.310 ","End":"01:16.510","Text":"We will just get back to the exercise."},{"Start":"01:16.510 ","End":"01:21.620","Text":"As I said, we\u0027ll break h up into f plus g under the condition that the absolute value of"},{"Start":"01:21.620 ","End":"01:27.665","Text":"g is less than the absolute value of f on this contour, Gamma_3."},{"Start":"01:27.665 ","End":"01:33.110","Text":"We want to do this in such a way that the number of zeros of f is easy to find."},{"Start":"01:33.110 ","End":"01:35.135","Text":"So we have to somehow break this h up."},{"Start":"01:35.135 ","End":"01:38.390","Text":"A bit of trial and error and we break h up as follows."},{"Start":"01:38.390 ","End":"01:41.745","Text":"We take f as the minus 5z to the fourth,"},{"Start":"01:41.745 ","End":"01:47.110","Text":"the middle term and g(z) will be 1/2z to the 6 plus 7z."},{"Start":"01:47.110 ","End":"01:49.715","Text":"Now we\u0027re going to verify this inequality."},{"Start":"01:49.715 ","End":"01:54.605","Text":"On Gamma_3, the absolute value of z is 3 and so"},{"Start":"01:54.605 ","End":"02:01.360","Text":"the absolute value of f(z) is the absolute value of minus 5z^4."},{"Start":"02:01.360 ","End":"02:05.120","Text":"We can write it this way and the absolute value of z is 3,"},{"Start":"02:05.120 ","End":"02:08.180","Text":"so it comes out to be 405."},{"Start":"02:08.180 ","End":"02:09.260","Text":"On the other hand,"},{"Start":"02:09.260 ","End":"02:11.425","Text":"the absolute value of g(z),"},{"Start":"02:11.425 ","End":"02:14.405","Text":"use the triangle inequality."},{"Start":"02:14.405 ","End":"02:17.180","Text":"You could follow this computation,"},{"Start":"02:17.180 ","End":"02:18.410","Text":"just pause the clip."},{"Start":"02:18.410 ","End":"02:21.140","Text":"It comes out to be 385.5."},{"Start":"02:21.140 ","End":"02:24.290","Text":"The point is that, that\u0027s less than 405."},{"Start":"02:24.290 ","End":"02:27.120","Text":"So this inequality is satisfied."},{"Start":"02:27.120 ","End":"02:30.225","Text":"We\u0027ve satisfied the conditions of Rouché\u0027s theorem."},{"Start":"02:30.225 ","End":"02:35.300","Text":"We have the conclusion that h and f have the same number of zeros."},{"Start":"02:35.300 ","End":"02:39.920","Text":"Let\u0027s see, the zeros of f are easy to compute because f is just minus 5z^4"},{"Start":"02:39.920 ","End":"02:46.150","Text":"and the only 0 in the disc or anywhere is z=0."},{"Start":"02:46.150 ","End":"02:53.270","Text":"This has an order 4 because of the z^4 here and so that means that the number of"},{"Start":"02:53.270 ","End":"03:00.260","Text":"zeros of h in the same disc is also 4, that\u0027s this part."},{"Start":"03:00.260 ","End":"03:03.765","Text":"Now we\u0027ll proceed to the other parts."},{"Start":"03:03.765 ","End":"03:10.485","Text":"Just to repeat, h has 4 zeros in this disc of radius 3."},{"Start":"03:10.485 ","End":"03:18.430","Text":"Gamma_1 is the border of this inner disc and it\u0027s just where absolute value of z=1."},{"Start":"03:18.430 ","End":"03:23.900","Text":"We want a different breakup of h into f plus g that will work for the inner disc."},{"Start":"03:23.900 ","End":"03:26.840","Text":"This time, again by trial and error,"},{"Start":"03:26.840 ","End":"03:36.770","Text":"we get that f will be the 7z part and g will be the 1.5 z^6 minus 5z."},{"Start":"03:36.770 ","End":"03:39.755","Text":"Let\u0027s do the computation for the inequalities."},{"Start":"03:39.755 ","End":"03:42.860","Text":"We said the absolute value of z=1 here."},{"Start":"03:42.860 ","End":"03:48.880","Text":"Absolute value of f(z) is absolute value of 7z comes out to be 7."},{"Start":"03:48.880 ","End":"03:51.705","Text":"Absolute value of g(z),"},{"Start":"03:51.705 ","End":"03:59.225","Text":"straightforward triangle inequality and absolute values comes out to be 5.5."},{"Start":"03:59.225 ","End":"04:04.010","Text":"The important thing is that this is less than the 7 here."},{"Start":"04:04.010 ","End":"04:11.015","Text":"We\u0027ve shown again that absolute value of g is less than absolute value of f on Gamma_1."},{"Start":"04:11.015 ","End":"04:14.120","Text":"The number of zeros of f is easy to compute."},{"Start":"04:14.120 ","End":"04:19.520","Text":"The number of zeros of 7z is just 1 everywhere in the complex plane,"},{"Start":"04:19.520 ","End":"04:21.620","Text":"in particular in this disc."},{"Start":"04:21.620 ","End":"04:24.890","Text":"It\u0027s just z=0, it\u0027s the only 0 and and it\u0027s order 1."},{"Start":"04:24.890 ","End":"04:26.960","Text":"That means by Rouché\u0027s theorem,"},{"Start":"04:26.960 ","End":"04:30.330","Text":"the number zeros of h is also equal to 1."},{"Start":"04:30.330 ","End":"04:33.450","Text":"So h has one 0 in this disc."},{"Start":"04:33.450 ","End":"04:35.160","Text":"That\u0027s the outer disc,"},{"Start":"04:35.160 ","End":"04:37.790","Text":"the inner disc, and now what about on the border?"},{"Start":"04:37.790 ","End":"04:41.810","Text":"Well, on Gamma_1 itself, absolute value of h,"},{"Start":"04:41.810 ","End":"04:47.450","Text":"which is absolute value of f plus g. This is bigger or equal to bivariant of"},{"Start":"04:47.450 ","End":"04:51.050","Text":"the triangle inequality to absolute value of f minus"},{"Start":"04:51.050 ","End":"04:55.415","Text":"absolute value of g and since this as an absolute value is smaller than this,"},{"Start":"04:55.415 ","End":"04:57.380","Text":"this is strictly positive."},{"Start":"04:57.380 ","End":"05:00.455","Text":"Positive in particular means not 0."},{"Start":"05:00.455 ","End":"05:08.975","Text":"So h is not equal to 0 on this circle and so the number of zeros is 0."},{"Start":"05:08.975 ","End":"05:12.470","Text":"That\u0027s all we need to finish the computation."},{"Start":"05:12.470 ","End":"05:18.410","Text":"We had that the number of zeros of h in D is equal to this minus this minus this."},{"Start":"05:18.410 ","End":"05:19.915","Text":"We showed that this is 4,"},{"Start":"05:19.915 ","End":"05:21.670","Text":"this is 0, and this is 1."},{"Start":"05:21.670 ","End":"05:28.900","Text":"So we get 4 minus 0 minus 1 and the answer is 3. We\u0027re done."}],"ID":23773},{"Watched":false,"Name":"Exercise 8","Duration":"3m 27s","ChapterTopicVideoID":22919,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.960","Text":"In this exercise, m is some integer,"},{"Start":"00:03.960 ","End":"00:06.450","Text":"I should say, bigger or equal to 2."},{"Start":"00:06.450 ","End":"00:13.515","Text":"Then, we have to find the number of zeros in the open unit disc of the function h,"},{"Start":"00:13.515 ","End":"00:21.660","Text":"which is defined as ze^m minus z minus 1 and this m here is the same as this m here."},{"Start":"00:21.660 ","End":"00:24.240","Text":"We\u0027re going to use Rouche\u0027s theorem of course."},{"Start":"00:24.240 ","End":"00:28.335","Text":"Let Gamma be the boundary of the unit circle."},{"Start":"00:28.335 ","End":"00:31.560","Text":"Sketch not much to show here,"},{"Start":"00:31.560 ","End":"00:33.345","Text":"but there it is."},{"Start":"00:33.345 ","End":"00:38.390","Text":"What we\u0027re going to do is break up h into f plus g,"},{"Start":"00:38.390 ","End":"00:41.840","Text":"such that the absolute value of g is less than"},{"Start":"00:41.840 ","End":"00:46.100","Text":"the absolute value of f on the boundary and"},{"Start":"00:46.100 ","End":"00:50.210","Text":"also so that it\u0027s easier to find the number of zeros of"},{"Start":"00:50.210 ","End":"00:54.770","Text":"f. Because Rouche\u0027s theorem says that h and f will have the same number of zeros."},{"Start":"00:54.770 ","End":"00:56.255","Text":"Now h is hard to find,"},{"Start":"00:56.255 ","End":"00:57.995","Text":"and hopefully f will be easy."},{"Start":"00:57.995 ","End":"01:02.900","Text":"Not many combinations to break up h and the one that works is to let"},{"Start":"01:02.900 ","End":"01:08.180","Text":"f be the first term and g is the second term,"},{"Start":"01:08.180 ","End":"01:14.690","Text":"so of course h is f plus g. Let\u0027s check that this inequality holds on Gamma."},{"Start":"01:14.690 ","End":"01:17.965","Text":"Here, the absolute value of z is 1,"},{"Start":"01:17.965 ","End":"01:22.460","Text":"so what we get is little computation absolute value of f,"},{"Start":"01:22.460 ","End":"01:25.520","Text":"which is absolute value of z^m-z."},{"Start":"01:25.520 ","End":"01:27.020","Text":"I think we can follow this anyway."},{"Start":"01:27.020 ","End":"01:36.305","Text":"We get to e^m over absolute value of e^z and because m is bigger or equal to 2 from here,"},{"Start":"01:36.305 ","End":"01:41.425","Text":"this is bigger or equal to e^2 over absolute value of e^z."},{"Start":"01:41.425 ","End":"01:43.730","Text":"I can\u0027t drop the absolute value here, by the way,"},{"Start":"01:43.730 ","End":"01:46.220","Text":"because the exponential is not always positive,"},{"Start":"01:46.220 ","End":"01:47.450","Text":"it\u0027s positive on the reals,"},{"Start":"01:47.450 ","End":"01:49.025","Text":"but in general not."},{"Start":"01:49.025 ","End":"01:55.265","Text":"This absolute value of e^z is equal to e to the power of the real part of z."},{"Start":"01:55.265 ","End":"02:01.500","Text":"Like you could write z equals x plus iy and the absolute value of e^iy is 1,"},{"Start":"02:01.500 ","End":"02:06.325","Text":"so we just get e^x or e to the power of real z and"},{"Start":"02:06.325 ","End":"02:11.960","Text":"that\u0027s less than or equal to e^1 because on this circle,"},{"Start":"02:11.960 ","End":"02:15.180","Text":"the real part of z is less than or equal to 1."},{"Start":"02:15.180 ","End":"02:19.240","Text":"One, is the maximum of the real part of z."},{"Start":"02:19.240 ","End":"02:21.230","Text":"What we get is that,"},{"Start":"02:21.230 ","End":"02:25.580","Text":"absolute value of f(z) is bigger or equal to e^2 over e^1."},{"Start":"02:25.580 ","End":"02:28.940","Text":"Notice the inequality changes sides here,"},{"Start":"02:28.940 ","End":"02:32.000","Text":"and this is equal to a."},{"Start":"02:32.000 ","End":"02:35.720","Text":"So we\u0027ve estimated absolute value of f, now what about g?"},{"Start":"02:35.720 ","End":"02:37.985","Text":"Absolute value of g,"},{"Start":"02:37.985 ","End":"02:39.950","Text":"g(z) is just the constant minus 1,"},{"Start":"02:39.950 ","End":"02:45.350","Text":"so it\u0027s equal to 1 and 1 is less than e and e is less than or"},{"Start":"02:45.350 ","End":"02:51.040","Text":"equal to f(z) so that we have that this is strictly less than this."},{"Start":"02:51.040 ","End":"02:53.160","Text":"That\u0027s what we wanted."},{"Start":"02:53.160 ","End":"02:57.244","Text":"So f and h have the same number of zeros."},{"Start":"02:57.244 ","End":"03:01.145","Text":"Now, the only 0 of f is at z equals 0,"},{"Start":"03:01.145 ","End":"03:03.650","Text":"e to the something is never zeros,"},{"Start":"03:03.650 ","End":"03:08.225","Text":"so it\u0027s only got a 0 when z is 0, and it\u0027s in order 1,0."},{"Start":"03:08.225 ","End":"03:10.745","Text":"So even counting multiplicity,"},{"Start":"03:10.745 ","End":"03:15.765","Text":"number of zeros of f is 1 inside the disk I should say."},{"Start":"03:15.765 ","End":"03:17.960","Text":"Now, but Rouche\u0027s theorem, as we said,"},{"Start":"03:17.960 ","End":"03:22.070","Text":"number of zeros of h is same as number of zeros of f. So h"},{"Start":"03:22.070 ","End":"03:27.750","Text":"also has exactly 1,0 in d and that\u0027s what we have to show, so we\u0027re done."}],"ID":23774},{"Watched":false,"Name":"Exercise 9","Duration":"9m 59s","ChapterTopicVideoID":22920,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.125","Text":"Here\u0027s another exercise where we\u0027ll use Rouche\u0027s theorem."},{"Start":"00:04.125 ","End":"00:08.130","Text":"Let D be the open unit disc."},{"Start":"00:08.130 ","End":"00:09.180","Text":"You could write it this way."},{"Start":"00:09.180 ","End":"00:12.390","Text":"It\u0027s also abbreviated D(0,1)."},{"Start":"00:12.390 ","End":"00:18.510","Text":"Suppose we have a_1 up to a_n in D, all different,"},{"Start":"00:18.510 ","End":"00:24.015","Text":"let\u0027s denote B(z) as the product from 1 to n"},{"Start":"00:24.015 ","End":"00:30.495","Text":"of z minus a_j over 1 minus z a_j conjugate,"},{"Start":"00:30.495 ","End":"00:33.619","Text":"where j is the running index."},{"Start":"00:33.619 ","End":"00:36.185","Text":"This is the Pi notation,"},{"Start":"00:36.185 ","End":"00:39.110","Text":"which is a bit like the Sigma notation,"},{"Start":"00:39.110 ","End":"00:42.710","Text":"only the Sigma relates to addition and Pi,"},{"Start":"00:42.710 ","End":"00:46.175","Text":"like p for product relates to multiplication."},{"Start":"00:46.175 ","End":"00:48.745","Text":"If we had pluses here, it would be Sigma,"},{"Start":"00:48.745 ","End":"00:52.800","Text":"multiplication Pie, but you\u0027ve probably seen it before."},{"Start":"00:52.800 ","End":"00:55.695","Text":"I\u0027ve written it out long handed."},{"Start":"00:55.695 ","End":"01:02.240","Text":"Let z naught be some point in D. We have to prove that the"},{"Start":"01:02.240 ","End":"01:10.355","Text":"function B(z) minus z naught has n roots in D, including multiplicities."},{"Start":"01:10.355 ","End":"01:17.520","Text":"We\u0027re given a hint to show that each of these factors in the product Phi_j,"},{"Start":"01:17.520 ","End":"01:20.250","Text":"we\u0027ll call it, which is z minus a_j over 1"},{"Start":"01:20.250 ","End":"01:25.065","Text":"minus z a_j conjugate maps the unit circle to itself."},{"Start":"01:25.065 ","End":"01:28.040","Text":"Before we properly start there\u0027s something that we"},{"Start":"01:28.040 ","End":"01:31.070","Text":"have to show to be technically correct."},{"Start":"01:31.070 ","End":"01:34.040","Text":"We have to show that this B(z) is well-defined."},{"Start":"01:34.040 ","End":"01:37.430","Text":"Basically, that the denominators here are not 0 because that\u0027s"},{"Start":"01:37.430 ","End":"01:41.045","Text":"the only impediment that there could be to defining it."},{"Start":"01:41.045 ","End":"01:50.160","Text":"We have to show that for each j from 1 to n,"},{"Start":"01:50.160 ","End":"01:55.350","Text":"that this denominator, 1 minus z a_j bar is not"},{"Start":"01:55.350 ","End":"02:03.210","Text":"0 for any z in D and we\u0027ll show this by contradiction."},{"Start":"02:03.210 ","End":"02:09.170","Text":"Supposing that 1 minus z a_j bar is 0, in that case,"},{"Start":"02:09.170 ","End":"02:10.985","Text":"we can extract z from here,"},{"Start":"02:10.985 ","End":"02:14.540","Text":"bring this to the other side and divide by a_j,"},{"Start":"02:14.960 ","End":"02:17.840","Text":"divide by a_j bar."},{"Start":"02:17.840 ","End":"02:23.525","Text":"Then we get that the modulus of z is 1 over the modulus of"},{"Start":"02:23.525 ","End":"02:31.230","Text":"a_j bar and the conjugate has the same absolute value as the number itself."},{"Start":"02:31.230 ","End":"02:33.200","Text":"So it\u0027s equal to this."},{"Start":"02:33.200 ","End":"02:35.510","Text":"Now, this is inside D,"},{"Start":"02:35.510 ","End":"02:37.269","Text":"which is the open unit circle."},{"Start":"02:37.269 ","End":"02:42.725","Text":"So the absolute value of the modulus is less than 1 and 1 over something less than 1,"},{"Start":"02:42.725 ","End":"02:45.800","Text":"positive is bigger than 1."},{"Start":"02:45.800 ","End":"02:48.845","Text":"There\u0027s one other thing I didn\u0027t write."},{"Start":"02:48.845 ","End":"02:52.640","Text":"You could say what if a_j was 0?"},{"Start":"02:52.640 ","End":"02:56.764","Text":"Well, if a_j is 0 and a_j bar is 0,"},{"Start":"02:56.764 ","End":"03:01.085","Text":"then this thing is definitely not 0."},{"Start":"03:01.085 ","End":"03:07.040","Text":"That applies automatically whenever a_j bar is 0."},{"Start":"03:07.040 ","End":"03:09.670","Text":"In either case, we\u0027re covered."},{"Start":"03:09.670 ","End":"03:14.020","Text":"I didn\u0027t write down everything this gets too technical."},{"Start":"03:14.090 ","End":"03:18.860","Text":"We\u0027ve just shown here that if this thing is 0,"},{"Start":"03:18.860 ","End":"03:21.830","Text":"then the absolute value of z is bigger than 1."},{"Start":"03:21.830 ","End":"03:23.300","Text":"On the other hand,"},{"Start":"03:23.300 ","End":"03:26.075","Text":"z is in the disk D,"},{"Start":"03:26.075 ","End":"03:30.170","Text":"so absolute value of z is less than 1 and that gives us"},{"Start":"03:30.170 ","End":"03:35.675","Text":"a contradiction and the contradiction shows that this can\u0027t be 0."},{"Start":"03:35.675 ","End":"03:42.020","Text":"Next, the hint will show that Phi sub j maps the unit circle to"},{"Start":"03:42.020 ","End":"03:49.485","Text":"itself and that means that Phi_j maps Gamma into Gamma."},{"Start":"03:49.485 ","End":"03:51.435","Text":"The image of Gamma under Phi_j,"},{"Start":"03:51.435 ","End":"03:53.390","Text":"or I guess I didn\u0027t say what Gamma is."},{"Start":"03:53.390 ","End":"03:54.905","Text":"Gamma is the unit circle."},{"Start":"03:54.905 ","End":"03:57.770","Text":"We want to show that Phi_j maps it to itself."},{"Start":"03:57.770 ","End":"04:01.430","Text":"In other words, that if absolute value of z is 1,"},{"Start":"04:01.430 ","End":"04:03.140","Text":"meaning z is on the unit circle,"},{"Start":"04:03.140 ","End":"04:09.220","Text":"then so is Phi_j of z and absolute value is 1 means on the unit circle."},{"Start":"04:09.220 ","End":"04:13.080","Text":"Let\u0027s evaluate this thing squared."},{"Start":"04:13.080 ","End":"04:14.650","Text":"If this is equal to 1,"},{"Start":"04:14.650 ","End":"04:17.135","Text":"then the absolute value is 1."},{"Start":"04:17.135 ","End":"04:18.500","Text":"This is equal 2,"},{"Start":"04:18.500 ","End":"04:21.200","Text":"by the definition of Phi_j here it is."},{"Start":"04:21.200 ","End":"04:27.080","Text":"Put it here, square it and then this is equal to applying the formula that"},{"Start":"04:27.080 ","End":"04:33.460","Text":"the modulus of something squared is the number times its conjugate."},{"Start":"04:33.460 ","End":"04:37.160","Text":"Then apply the conjugate."},{"Start":"04:37.160 ","End":"04:41.180","Text":"Conjugate applies to sums, differences, and quotients."},{"Start":"04:41.180 ","End":"04:42.980","Text":"We take the conjugate of the numerator,"},{"Start":"04:42.980 ","End":"04:44.735","Text":"conjugate of the denominator,"},{"Start":"04:44.735 ","End":"04:49.400","Text":"and also the conjugate of the conjugate is the number itself and we get to this,"},{"Start":"04:49.400 ","End":"04:53.450","Text":"and then we can just multiply out top and bottom like"},{"Start":"04:53.450 ","End":"04:59.295","Text":"a fraction z times z bar is absolute value of z^2."},{"Start":"04:59.295 ","End":"05:01.790","Text":"I accidentally left the blank space here."},{"Start":"05:01.790 ","End":"05:04.040","Text":"Anyway, as I was saying, a number times,"},{"Start":"05:04.040 ","End":"05:09.700","Text":"its conjugate is the modulus squared and we have that several times."},{"Start":"05:09.700 ","End":"05:11.920","Text":"We also have it when we multiply this term,"},{"Start":"05:11.920 ","End":"05:15.280","Text":"we have this z with this z bar,"},{"Start":"05:15.280 ","End":"05:19.555","Text":"z conjugate is equal to modulus of z^2."},{"Start":"05:19.555 ","End":"05:23.905","Text":"We also have that a_j times"},{"Start":"05:23.905 ","End":"05:30.870","Text":"a_j conjugate is a_j absolute value squared."},{"Start":"05:30.870 ","End":"05:38.460","Text":"Also down here with this gives us the modulus squared."},{"Start":"05:38.460 ","End":"05:41.710","Text":"Although this looks like a complicated expression,"},{"Start":"05:41.710 ","End":"05:46.540","Text":"I claim it\u0027s exactly equal to 1 on this unit circle."},{"Start":"05:46.540 ","End":"05:48.850","Text":"This is because every term in"},{"Start":"05:48.850 ","End":"05:53.330","Text":"the numerator equals the corresponding term in the denominator."},{"Start":"05:53.330 ","End":"05:57.470","Text":"Look, the modulus of z^2 we already said,"},{"Start":"05:57.470 ","End":"06:01.440","Text":"is equal to 1 because of this."},{"Start":"06:01.440 ","End":"06:06.015","Text":"Now, this and this are just the same."},{"Start":"06:06.015 ","End":"06:12.735","Text":"Also this and this are the same."},{"Start":"06:12.735 ","End":"06:16.850","Text":"Here also because the modulus of z is 1,"},{"Start":"06:16.850 ","End":"06:19.910","Text":"this is the same as itself times 1."},{"Start":"06:19.910 ","End":"06:21.170","Text":"I\u0027m not canceling here,"},{"Start":"06:21.170 ","End":"06:23.690","Text":"like usual I\u0027m just checking off each pair."},{"Start":"06:23.690 ","End":"06:26.210","Text":"This is the same as this, this is the same as this and so on."},{"Start":"06:26.210 ","End":"06:30.170","Text":"Everything getting upstairs is the same as what we have downstairs,"},{"Start":"06:30.170 ","End":"06:33.355","Text":"so the quotient is 1."},{"Start":"06:33.355 ","End":"06:35.450","Text":"On the unit circle,"},{"Start":"06:35.450 ","End":"06:39.980","Text":"we have that the modulus of B(z) which"},{"Start":"06:39.980 ","End":"06:45.580","Text":"is the product of this that\u0027s just by the definition, just putting bars."},{"Start":"06:45.580 ","End":"06:49.925","Text":"Now, each of these is equal to 1 and we\u0027ve just shown."},{"Start":"06:49.925 ","End":"06:55.350","Text":"What we have is the product of n factors each of them is 1."},{"Start":"06:55.350 ","End":"06:56.670","Text":"So 1 times 1 times 1,"},{"Start":"06:56.670 ","End":"06:58.435","Text":"n times is 1."},{"Start":"06:58.435 ","End":"07:04.670","Text":"We\u0027ve actually shown that not only do the individual Phi_j map the unit circle to itself,"},{"Start":"07:04.670 ","End":"07:07.880","Text":"but the function capital B takes"},{"Start":"07:07.880 ","End":"07:11.300","Text":"the unit circle to itself because if absolute value of z is 1,"},{"Start":"07:11.300 ","End":"07:13.910","Text":"absolute value of B(z) is also 1."},{"Start":"07:13.910 ","End":"07:15.706","Text":"Now, some definitions."},{"Start":"07:15.706 ","End":"07:19.885","Text":"We\u0027ll define 3 functions we\u0027ll define h,"},{"Start":"07:19.885 ","End":"07:25.145","Text":"we\u0027ll define f and we\u0027ll define g just as written here."},{"Start":"07:25.145 ","End":"07:28.125","Text":"The idea is we\u0027re going to use Rouche\u0027s theorem of course."},{"Start":"07:28.125 ","End":"07:30.350","Text":"Notice that h is equal to f plus"},{"Start":"07:30.350 ","End":"07:35.520","Text":"g. B(z) plus minus z naught gives us B(z) minus z naught,"},{"Start":"07:35.520 ","End":"07:36.930","Text":"which is h. Now,"},{"Start":"07:36.930 ","End":"07:40.370","Text":"what we\u0027re going to do is show that the absolute value of"},{"Start":"07:40.370 ","End":"07:44.045","Text":"f is bigger than the absolute value of g on Gamma,"},{"Start":"07:44.045 ","End":"07:46.865","Text":"the unit circle and then we\u0027ll apply Rouche\u0027s theorem."},{"Start":"07:46.865 ","End":"07:49.010","Text":"First of all, the left-hand side."},{"Start":"07:49.010 ","End":"07:50.930","Text":"If z is on Gamma,"},{"Start":"07:50.930 ","End":"07:56.120","Text":"then absolute value of z is 1 and so absolute value of B(z) is 1,"},{"Start":"07:56.120 ","End":"08:02.375","Text":"which we just showed B is f. Absolute value of f(z) is 1."},{"Start":"08:02.375 ","End":"08:04.210","Text":"Now, the other one, g(z)."},{"Start":"08:04.210 ","End":"08:08.450","Text":"Absolute value of g(z) is the same as the absolute value of"},{"Start":"08:08.450 ","End":"08:13.850","Text":"z naught and that\u0027s less than 1 since z naught is in the open unit disc."},{"Start":"08:13.850 ","End":"08:16.520","Text":"Something less than 1 and something equal"},{"Start":"08:16.520 ","End":"08:19.640","Text":"1 so absolute value of g is less than"},{"Start":"08:19.640 ","End":"08:22.880","Text":"absolute value of f strictly less on the unit circle."},{"Start":"08:22.880 ","End":"08:26.935","Text":"That means what we fulfill the conditions of Rouche\u0027s theorem."},{"Start":"08:26.935 ","End":"08:30.470","Text":"That means that h and f have the same number of"},{"Start":"08:30.470 ","End":"08:35.100","Text":"zeros inside of D. In other words, what is h?"},{"Start":"08:35.100 ","End":"08:38.555","Text":"B(z) minus z naught has the same number of zeros as f,"},{"Start":"08:38.555 ","End":"08:43.100","Text":"which is B(z) in D. What we have to do now is show"},{"Start":"08:43.100 ","End":"08:47.930","Text":"that this number is n. The last thing we have to do,"},{"Start":"08:47.930 ","End":"08:56.380","Text":"and then we\u0027re done is to show that B(z) has n zeros in D counting multiplicity."},{"Start":"08:56.380 ","End":"09:03.260","Text":"Well, the zeros of B(z) which is this product,"},{"Start":"09:03.260 ","End":"09:05.660","Text":"are clearly just the a_j."},{"Start":"09:05.660 ","End":"09:09.080","Text":"A quotient is 0 if and only if the numerator is"},{"Start":"09:09.080 ","End":"09:13.750","Text":"0 and the product is 0 if and only if at least one of the factors is 0."},{"Start":"09:13.750 ","End":"09:19.310","Text":"The only way this can be 0 is at z=a_j for some j."},{"Start":"09:19.310 ","End":"09:20.960","Text":"These are the zeros."},{"Start":"09:20.960 ","End":"09:23.855","Text":"Now, I\u0027ll just point out that it doesn\u0027t really matter"},{"Start":"09:23.855 ","End":"09:27.650","Text":"that the a_j are all different as we were told."},{"Start":"09:27.650 ","End":"09:29.975","Text":"If a_j appear twice,"},{"Start":"09:29.975 ","End":"09:36.885","Text":"then we just get a 0 of order 2 and it would still be n zeros each one contributes 1."},{"Start":"09:36.885 ","End":"09:39.800","Text":"I guess it was superfluous to require"},{"Start":"09:39.800 ","End":"09:42.910","Text":"that these will be different anyways. There are n of them."},{"Start":"09:42.910 ","End":"09:47.720","Text":"J from 1 to n and they\u0027re all inside D,"},{"Start":"09:47.720 ","End":"09:52.630","Text":"because we were given that the a_j were inside D. That\u0027s it."},{"Start":"09:52.630 ","End":"09:56.390","Text":"We\u0027ve shown that B(z) has n zeros inside D,"},{"Start":"09:56.390 ","End":"10:00.270","Text":"and that was the missing piece and we are done."}],"ID":23775},{"Watched":false,"Name":"Exercise 10","Duration":"6m 32s","ChapterTopicVideoID":22921,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.650","Text":"In this exercise, we have to show that the equation sine z equals z to the n has"},{"Start":"00:07.650 ","End":"00:15.720","Text":"exactly one solution in the disk with radius a 1/2,"},{"Start":"00:15.720 ","End":"00:18.075","Text":"but not for every n,"},{"Start":"00:18.075 ","End":"00:22.455","Text":"just from a certain number n onwards."},{"Start":"00:22.455 ","End":"00:30.035","Text":"There is such an N, that for all subsequent values of little n, this equation holds."},{"Start":"00:30.035 ","End":"00:32.760","Text":"We\u0027ll be using Rouche\u0027s theorem."},{"Start":"00:32.760 ","End":"00:36.090","Text":"Just to remind you what it is,"},{"Start":"00:36.090 ","End":"00:41.450","Text":"you should pause this video if necessary and remind yourself."},{"Start":"00:41.450 ","End":"00:43.505","Text":"Anyway, let\u0027s get started."},{"Start":"00:43.505 ","End":"00:47.945","Text":"Now, the solutions of sin (z)= z^n are"},{"Start":"00:47.945 ","End":"00:54.260","Text":"the roots of the function sin (z) minus z^n,"},{"Start":"00:54.260 ","End":"00:55.825","Text":"when this equals this,"},{"Start":"00:55.825 ","End":"01:01.040","Text":"then this minus this is 0 so we\u0027re talking about the roots."},{"Start":"01:01.040 ","End":"01:03.665","Text":"Like I said, we use Rouche\u0027s theorem."},{"Start":"01:03.665 ","End":"01:09.055","Text":"We want to break h up into f plus g,"},{"Start":"01:09.055 ","End":"01:12.470","Text":"and for all z on Gamma,"},{"Start":"01:12.470 ","End":"01:17.135","Text":"which is the boundary of this as Gamma here,"},{"Start":"01:17.135 ","End":"01:19.040","Text":"absolute value of g is less than,"},{"Start":"01:19.040 ","End":"01:22.040","Text":"absolute value of f strictly less than."},{"Start":"01:22.040 ","End":"01:29.930","Text":"We also want this breakup to be such that f is easier to figure out the roots"},{"Start":"01:29.930 ","End":"01:34.130","Text":"of than h. There\u0027s two obvious choices of f"},{"Start":"01:34.130 ","End":"01:38.510","Text":"and g either we let the first part be f and the second part be g,"},{"Start":"01:38.510 ","End":"01:42.060","Text":"or this is g and this is f,"},{"Start":"01:42.060 ","End":"01:44.810","Text":"one or the other will probably be the right thing,"},{"Start":"01:44.810 ","End":"01:46.685","Text":"and which is going to be?"},{"Start":"01:46.685 ","End":"01:55.965","Text":"Well, informally I\u0027ll say that the absolute value of sine z is bounded away from 0,"},{"Start":"01:55.965 ","End":"01:58.110","Text":"it doesn\u0027t get near 0."},{"Start":"01:58.110 ","End":"01:59.400","Text":"But the other one,"},{"Start":"01:59.400 ","End":"02:00.975","Text":"z to the n,"},{"Start":"02:00.975 ","End":"02:02.415","Text":"minus z to the n,"},{"Start":"02:02.415 ","End":"02:05.470","Text":"tends to 0 uniformly and evenly."},{"Start":"02:05.470 ","End":"02:08.675","Text":"You don\u0027t have to know that term, won\u0027t use that."},{"Start":"02:08.675 ","End":"02:14.620","Text":"Anyway, so what it means is that the second one is smaller than the first one."},{"Start":"02:14.620 ","End":"02:17.120","Text":"The absolute value of minus z to the n is"},{"Start":"02:17.120 ","End":"02:19.685","Text":"smaller than absolute value of sine z for large n,"},{"Start":"02:19.685 ","End":"02:24.845","Text":"so we\u0027re going to choose this to be f and this to be g,"},{"Start":"02:24.845 ","End":"02:28.210","Text":"g the smaller one because of this."},{"Start":"02:28.210 ","End":"02:33.875","Text":"Let\u0027s just make this more precise by the extreme value theorem,"},{"Start":"02:33.875 ","End":"02:41.540","Text":"if we let m be the minimum of sine z on Gamma because Gamma is compact,"},{"Start":"02:41.540 ","End":"02:43.075","Text":"it\u0027s closed and bounded,"},{"Start":"02:43.075 ","End":"02:45.920","Text":"then there is a minimum and the minimum is attained at"},{"Start":"02:45.920 ","End":"02:50.140","Text":"some point on this boundary, call it z_min."},{"Start":"02:50.140 ","End":"02:53.940","Text":"Let\u0027s say here somewhere on the boundaries and I claim that"},{"Start":"02:53.940 ","End":"02:59.300","Text":"the sine at this point z minimum is not 0."},{"Start":"02:59.300 ","End":"03:00.830","Text":"That will do is by contradiction."},{"Start":"03:00.830 ","End":"03:04.105","Text":"Suppose sine of z_min is 0."},{"Start":"03:04.105 ","End":"03:07.640","Text":"Then z_min is one of the solutions to sine z"},{"Start":"03:07.640 ","End":"03:11.225","Text":"equals 0 and the solutions are just these values,"},{"Start":"03:11.225 ","End":"03:13.690","Text":"multiples of Pi,"},{"Start":"03:13.690 ","End":"03:19.654","Text":"so absolute value of z_min just throughout the plus or minus has to be one of these."},{"Start":"03:19.654 ","End":"03:23.690","Text":"It can\u0027t be a 1/2. A 1/2 is not in this set,"},{"Start":"03:23.690 ","End":"03:32.040","Text":"so z_min does not belong to Gamma because Gamma is where z_min is equal to a 1/2."},{"Start":"03:32.040 ","End":"03:33.780","Text":"We have z_min is not in Gamma,"},{"Start":"03:33.780 ","End":"03:35.865","Text":"but z_min is in Gamma,"},{"Start":"03:35.865 ","End":"03:38.730","Text":"and so that gives us the contradiction."},{"Start":"03:38.730 ","End":"03:42.140","Text":"The contradiction came from assuming this,"},{"Start":"03:42.140 ","End":"03:43.925","Text":"so putting this together,"},{"Start":"03:43.925 ","End":"03:45.935","Text":"for every z in Gamma,"},{"Start":"03:45.935 ","End":"03:52.415","Text":"the absolute value of sine z is bigger or equal to the minimum possible sine z,"},{"Start":"03:52.415 ","End":"03:59.600","Text":"which we call the m, is bigger than 0 because this minimum is sine of z_min."},{"Start":"03:59.600 ","End":"04:02.450","Text":"The absolute value of f of z,"},{"Start":"04:02.450 ","End":"04:04.070","Text":"which is sine z,"},{"Start":"04:04.070 ","End":"04:06.260","Text":"is bigger or equal to m,"},{"Start":"04:06.260 ","End":"04:10.515","Text":"which is to the positive for all z in Gamma."},{"Start":"04:10.515 ","End":"04:17.260","Text":"Now, what about g? Show that g(z) is less than m on Gamma and then will be settled with"},{"Start":"04:17.260 ","End":"04:24.580","Text":"the inequality that absolute value of g is less than absolute value of f on Gamma."},{"Start":"04:24.580 ","End":"04:29.260","Text":"Recall that g(z) is minus z to"},{"Start":"04:29.260 ","End":"04:34.180","Text":"the n. Absolute value of g(z) absolute value of minus z to the n,"},{"Start":"04:34.180 ","End":"04:35.560","Text":"the minus doesn\u0027t matter."},{"Start":"04:35.560 ","End":"04:38.710","Text":"The absolute value of z is a 1/2 on Gamma."},{"Start":"04:38.710 ","End":"04:42.940","Text":"This is 1/2 to the n and this goes to 0 as n goes to infinity,"},{"Start":"04:42.940 ","End":"04:46.835","Text":"which means that from some big n onwards,"},{"Start":"04:46.835 ","End":"04:51.385","Text":"1/2 to the power of little n is less than m,"},{"Start":"04:51.385 ","End":"04:53.830","Text":"where m is that minimum."},{"Start":"04:53.830 ","End":"04:55.630","Text":"We define this here,"},{"Start":"04:55.630 ","End":"04:59.150","Text":"m is the minimum and it was positive."},{"Start":"04:59.150 ","End":"05:03.755","Text":"Absolute value of g(z) is less than m,"},{"Start":"05:03.755 ","End":"05:06.580","Text":"and absolute value of f is bigger than or equal to m,"},{"Start":"05:06.580 ","End":"05:11.510","Text":"so that means that absolute value of g is less"},{"Start":"05:11.510 ","End":"05:16.520","Text":"than absolute value of f on Gamma and we filled"},{"Start":"05:16.520 ","End":"05:24.140","Text":"the conditions for Rouche\u0027s theorem which says that our original h has the same number of"},{"Start":"05:24.140 ","End":"05:28.730","Text":"zeros as f. This function sine z minus"},{"Start":"05:28.730 ","End":"05:33.440","Text":"z to the n are the same number of zeros or sine z in D,"},{"Start":"05:33.440 ","End":"05:39.635","Text":"and this is going to be easier to count because sine z has just 1 zero in D,"},{"Start":"05:39.635 ","End":"05:41.465","Text":"that\u0027s z equals 0."},{"Start":"05:41.465 ","End":"05:47.990","Text":"We look at the solutions for sine z it\u0027s multiples of Pi and the only multiple of Pi that"},{"Start":"05:47.990 ","End":"05:54.865","Text":"fits in the disk of radius a 1/2 is 0 because Pi is bigger than a 1/2."},{"Start":"05:54.865 ","End":"06:03.005","Text":"Now it\u0027s a simple 0 or 0 of multiplicity 1 because its derivative is not 0."},{"Start":"06:03.005 ","End":"06:09.155","Text":"Derivative of sine z is cosine z and when z is 0, that\u0027s not 0."},{"Start":"06:09.155 ","End":"06:12.485","Text":"Zero is a simple 0,"},{"Start":"06:12.485 ","End":"06:17.330","Text":"that means that this function h(z) has exactly one"},{"Start":"06:17.330 ","End":"06:22.880","Text":"zero in D because it has the same number of zeros or sine z and that\u0027s one,"},{"Start":"06:22.880 ","End":"06:24.305","Text":"so this is one,"},{"Start":"06:24.305 ","End":"06:33.750","Text":"and that means that this equation has exactly one solution in D and we\u0027re done."}],"ID":23776},{"Watched":false,"Name":"Exercise 11","Duration":"2m 43s","ChapterTopicVideoID":22922,"CourseChapterTopicPlaylistID":102292,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.750","Text":"In this exercise, we have a function f which is analytic on the closed unit disk and"},{"Start":"00:06.750 ","End":"00:14.430","Text":"satisfies absolute value of f is less than 1 on the unit circle,"},{"Start":"00:14.430 ","End":"00:17.535","Text":"which is the boundary of the unit disk."},{"Start":"00:17.535 ","End":"00:23.910","Text":"We have to prove that f has exactly 1 stationary point in the open unit disk."},{"Start":"00:23.910 ","End":"00:27.420","Text":"In case you don\u0027t know what a stationary point is, I\u0027ll rephrase that."},{"Start":"00:27.420 ","End":"00:31.290","Text":"In other words, we have to prove that the equation f of z equals"},{"Start":"00:31.290 ","End":"00:37.260","Text":"z has exactly 1 solution in the open unit disk."},{"Start":"00:37.260 ","End":"00:42.230","Text":"The plan is to show that f of z minus z as a function,"},{"Start":"00:42.230 ","End":"00:44.070","Text":"has exactly 1,"},{"Start":"00:44.070 ","End":"00:46.830","Text":"0 in the open unit disk,"},{"Start":"00:46.830 ","End":"00:52.290","Text":"and then it can only have 1 solution because if f of z minus z is 0,"},{"Start":"00:52.290 ","End":"00:54.240","Text":"then f of z equals z."},{"Start":"00:54.240 ","End":"00:56.910","Text":"We\u0027re going to use Rouche\u0027s theorem."},{"Start":"00:56.910 ","End":"01:00.650","Text":"I\u0027ll remind you, here it is."},{"Start":"01:00.650 ","End":"01:01.790","Text":"I won\u0027t linger on it."},{"Start":"01:01.790 ","End":"01:03.620","Text":"You can pause and review it,"},{"Start":"01:03.620 ","End":"01:05.960","Text":"let\u0027s get back to the exercise."},{"Start":"01:05.960 ","End":"01:08.030","Text":"Yeah, Rouche\u0027s theorem."},{"Start":"01:08.030 ","End":"01:10.490","Text":"I\u0027ll use capital letters F,"},{"Start":"01:10.490 ","End":"01:13.490","Text":"G, and H because little f is already taken."},{"Start":"01:13.490 ","End":"01:18.028","Text":"We\u0027ll let H be f of z minus z,"},{"Start":"01:18.028 ","End":"01:25.550","Text":"and we\u0027ll break it up as f of z is minus z and G of z is f of z,"},{"Start":"01:25.550 ","End":"01:28.610","Text":"so that H is F+G."},{"Start":"01:28.610 ","End":"01:32.975","Text":"Then Gamma, the boundary of D is the unit circle."},{"Start":"01:32.975 ","End":"01:40.040","Text":"We\u0027re going to show that absolute value of G is less than absolute value of F on Gamma."},{"Start":"01:40.040 ","End":"01:43.310","Text":"The absolute value of G of z is by definition,"},{"Start":"01:43.310 ","End":"01:46.115","Text":"absolute value of f of z,"},{"Start":"01:46.115 ","End":"01:52.105","Text":"and that\u0027s less than 1 on Gamma, that\u0027s given here."},{"Start":"01:52.105 ","End":"01:54.800","Text":"Absolute value of F of z,"},{"Start":"01:54.800 ","End":"01:58.490","Text":"which is absolute value of minus z is exactly 1"},{"Start":"01:58.490 ","End":"02:02.600","Text":"on Gamma because Gamma is where absolute value of z equals 1,"},{"Start":"02:02.600 ","End":"02:04.690","Text":"minus you can just throw it out."},{"Start":"02:04.690 ","End":"02:06.560","Text":"We have something that\u0027s equal to 1,"},{"Start":"02:06.560 ","End":"02:11.405","Text":"something is less than 1 so this is less than this on Gamma."},{"Start":"02:11.405 ","End":"02:17.245","Text":"By the theorem, H and F have the same number of zeros in D,"},{"Start":"02:17.245 ","End":"02:20.370","Text":"but F of z has exactly one 0 in D,"},{"Start":"02:20.370 ","End":"02:24.090","Text":"one 0 in low plane,"},{"Start":"02:24.090 ","End":"02:26.799","Text":"which is z equals 0."},{"Start":"02:27.140 ","End":"02:33.275","Text":"H also has exactly one 0 in D, and therefore,"},{"Start":"02:33.275 ","End":"02:37.600","Text":"f of z equals z has exactly 1 solution,"},{"Start":"02:37.600 ","End":"02:43.580","Text":"which is the 0 of H. That\u0027s it."}],"ID":23777}],"Thumbnail":null,"ID":102292},{"Name":"Hurwitz s Theorem","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Hurwitz\u0027s Theorem (with proof)","Duration":"7m 54s","ChapterTopicVideoID":22927,"CourseChapterTopicPlaylistID":102293,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"Now we come to a new topic."},{"Start":"00:03.045 ","End":"00:07.035","Text":"A theorem due to Hurwitz."},{"Start":"00:07.035 ","End":"00:09.930","Text":"Picture of him here."},{"Start":"00:09.930 ","End":"00:12.450","Text":"What he says is as follows."},{"Start":"00:12.450 ","End":"00:17.130","Text":"Suppose we have a sequence f_n of analytic functions on"},{"Start":"00:17.130 ","End":"00:24.075","Text":"a domain D in the complex plane and domain means open and connected,"},{"Start":"00:24.075 ","End":"00:30.900","Text":"which converges compactly on D and I\u0027ll remind you what that means."},{"Start":"00:30.900 ","End":"00:33.765","Text":"It means that it converges uniformly,"},{"Start":"00:33.765 ","End":"00:35.370","Text":"but not everywhere on D,"},{"Start":"00:35.370 ","End":"00:38.595","Text":"uniformly on each compact subset."},{"Start":"00:38.595 ","End":"00:40.590","Text":"I\u0027m going to put the picture here already,"},{"Start":"00:40.590 ","End":"00:41.865","Text":"so we\u0027ll have it."},{"Start":"00:41.865 ","End":"00:45.825","Text":"Meanwhile, there\u0027s D, the domain."},{"Start":"00:45.825 ","End":"00:50.360","Text":"The limit of this convergence is an analytic function."},{"Start":"00:50.360 ","End":"00:52.730","Text":"We don\u0027t have to require it to be analytic."},{"Start":"00:52.730 ","End":"00:58.580","Text":"It\u0027s necessarily analytic because of a theorem due to Weierstrass that"},{"Start":"00:58.580 ","End":"01:04.895","Text":"if you have a sequence f_n of analytic functions and f_n converges to f compactly,"},{"Start":"01:04.895 ","End":"01:08.870","Text":"then the limit f is also analytic."},{"Start":"01:08.870 ","End":"01:11.555","Text":"That\u0027s here so far."},{"Start":"01:11.555 ","End":"01:15.200","Text":"Now, suppose we have z_0 in D,"},{"Start":"01:15.200 ","End":"01:17.120","Text":"and that\u0027s this z_0 here,"},{"Start":"01:17.120 ","End":"01:22.830","Text":"and it\u0027s a 0 of f of order m. Notice that m is bigger or equal to 0."},{"Start":"01:22.830 ","End":"01:25.700","Text":"It could be a 0 of order 0,"},{"Start":"01:25.700 ","End":"01:27.755","Text":"meaning it\u0027s not a 0 at all."},{"Start":"01:27.755 ","End":"01:30.905","Text":"Rho, which is here,"},{"Start":"01:30.905 ","End":"01:37.070","Text":"is a positive real number such that the open disk centered at z_0 with"},{"Start":"01:37.070 ","End":"01:43.070","Text":"radius Rho is completely contained in D. That\u0027s this circle here."},{"Start":"01:43.070 ","End":"01:45.860","Text":"It\u0027s completely contained inside D,"},{"Start":"01:45.860 ","End":"01:53.630","Text":"and it has no zeros in it, except possibly z_0."},{"Start":"01:53.630 ","End":"01:57.650","Text":"That would depend on whether m is 0 or non-zero."},{"Start":"01:57.650 ","End":"02:04.010","Text":"Then for any r which is less than or equal to Rho also positive,"},{"Start":"02:04.010 ","End":"02:08.330","Text":"that would be smaller circle or it could be equal to."},{"Start":"02:08.330 ","End":"02:12.915","Text":"There exists some natural number N,"},{"Start":"02:12.915 ","End":"02:14.310","Text":"depending on r,"},{"Start":"02:14.310 ","End":"02:17.985","Text":"such that for little n bigger than this N,"},{"Start":"02:17.985 ","End":"02:22.080","Text":"f_n has exactly m 0s,"},{"Start":"02:22.080 ","End":"02:24.690","Text":"counting them with multiplicity,"},{"Start":"02:24.690 ","End":"02:31.100","Text":"in the open disk with radius little r. Let me explain this again."},{"Start":"02:31.100 ","End":"02:35.015","Text":"We have a domain D and we have z_0,"},{"Start":"02:35.015 ","End":"02:38.045","Text":"which is a 0 of order m,"},{"Start":"02:38.045 ","End":"02:40.895","Text":"and we have a circle around it,"},{"Start":"02:40.895 ","End":"02:44.775","Text":"which isolates it from any other possible 0s."},{"Start":"02:44.775 ","End":"02:52.920","Text":"Then the members of the sequence have m 0s inside this disk."},{"Start":"02:52.920 ","End":"02:54.470","Text":"However small it is,"},{"Start":"02:54.470 ","End":"02:59.480","Text":"but this is only true for big enough N from a certain point onwards,"},{"Start":"02:59.480 ","End":"03:06.170","Text":"all the m zeros of members of the sequence are inside small circle around z_0."},{"Start":"03:06.170 ","End":"03:09.410","Text":"Now before we go on to the proof of the theorem,"},{"Start":"03:09.410 ","End":"03:12.650","Text":"I just like to mention that there are"},{"Start":"03:12.650 ","End":"03:16.340","Text":"several variations in different textbooks or"},{"Start":"03:16.340 ","End":"03:20.895","Text":"different web pages on what Hurwitz theorem is,"},{"Start":"03:20.895 ","End":"03:25.120","Text":"in case you look it up and see that it\u0027s somewhat different."},{"Start":"03:25.120 ","End":"03:27.345","Text":"Now, the proof."},{"Start":"03:27.345 ","End":"03:31.740","Text":"Let\u0027s suppose that r is as here and"},{"Start":"03:31.740 ","End":"03:36.620","Text":"let Gamma r be the circle of radius r centered at z_0."},{"Start":"03:36.620 ","End":"03:39.670","Text":"This is this circle here."},{"Start":"03:39.670 ","End":"03:46.370","Text":"Now, Gamma r is completely contained inside the open disk of radius Rho,"},{"Start":"03:46.370 ","End":"03:49.100","Text":"where f has no 0s."},{"Start":"03:49.100 ","End":"03:53.945","Text":"It says here that f has no 0s in this open disk,"},{"Start":"03:53.945 ","End":"03:55.760","Text":"except possibly at z_0,"},{"Start":"03:55.760 ","End":"03:58.520","Text":"which is certainly not on this circle either."},{"Start":"03:58.520 ","End":"04:04.400","Text":"The absolute value of f is also continuous and it\u0027s real valued on"},{"Start":"04:04.400 ","End":"04:12.245","Text":"this compact set Gamma r. We will apply the extreme value theorem and by this theorem,"},{"Start":"04:12.245 ","End":"04:16.370","Text":"the absolute value of f achieves its minimum and the minimum"},{"Start":"04:16.370 ","End":"04:21.065","Text":"occurs at some point on this circle and it\u0027s called that z_min."},{"Start":"04:21.065 ","End":"04:23.030","Text":"Now it\u0027s the smallest,"},{"Start":"04:23.030 ","End":"04:26.600","Text":"so for any other z on the circle,"},{"Start":"04:26.600 ","End":"04:32.155","Text":"Gamma r, absolute value of f(z) is bigger or equal to absolute value of f(z_min)."},{"Start":"04:32.155 ","End":"04:39.900","Text":"This is bigger than 0 because f(z_min) is not equal to 0 for all z on Gamma r,"},{"Start":"04:39.900 ","End":"04:41.355","Text":"f is non zero."},{"Start":"04:41.355 ","End":"04:43.400","Text":"If this is bigger than 0,"},{"Start":"04:43.400 ","End":"04:47.675","Text":"then we can choose any Epsilon between 0 and it."},{"Start":"04:47.675 ","End":"04:51.440","Text":"You could say Epsilon equals 1/2 of this or whatever."},{"Start":"04:51.440 ","End":"04:57.560","Text":"Now, f_n converges to f uniformly on Gamma r because it converges uniformly on"},{"Start":"04:57.560 ","End":"04:59.780","Text":"each compact subset and in particular on"},{"Start":"04:59.780 ","End":"05:03.560","Text":"Gamma r. By the definition of uniformly converges,"},{"Start":"05:03.560 ","End":"05:10.595","Text":"there is some natural number N such that for all n bigger than this N,"},{"Start":"05:10.595 ","End":"05:16.220","Text":"the distance between f_n(z) and f(z) is less than Epsilon for"},{"Start":"05:16.220 ","End":"05:22.160","Text":"all z in Gamma r. It\u0027s the same n for all z."},{"Start":"05:22.160 ","End":"05:23.825","Text":"That\u0027s the uniform."},{"Start":"05:23.825 ","End":"05:30.050","Text":"Now let\u0027s define functions g_n for each n bigger than N_r."},{"Start":"05:30.050 ","End":"05:33.705","Text":"We\u0027ll let g_n be this difference."},{"Start":"05:33.705 ","End":"05:35.400","Text":"We\u0027re going to use Rouche\u0027s theorem,"},{"Start":"05:35.400 ","End":"05:37.200","Text":"so I better remind you."},{"Start":"05:37.200 ","End":"05:39.505","Text":"Here\u0027s Rouche\u0027s theorem."},{"Start":"05:39.505 ","End":"05:41.570","Text":"You can pause and study this."},{"Start":"05:41.570 ","End":"05:46.130","Text":"I\u0027ll just say briefly that if we have a function f,"},{"Start":"05:46.130 ","End":"05:53.120","Text":"which has a certain number of 0s in a domain D and g is smaller than f,"},{"Start":"05:53.120 ","End":"05:55.385","Text":"an absolute value on the boundary."},{"Start":"05:55.385 ","End":"05:57.640","Text":"Then if we add g to f,"},{"Start":"05:57.640 ","End":"06:04.300","Text":"the sum has the same number of 0s in D as the original f. Rouche\u0027s theorem."},{"Start":"06:04.300 ","End":"06:12.530","Text":"The function f will be just as is and g_n is what we called g in the theorem."},{"Start":"06:12.530 ","End":"06:14.990","Text":"F_n will be f plus g,"},{"Start":"06:14.990 ","End":"06:18.590","Text":"which in the exercises we called h. In other words,"},{"Start":"06:18.590 ","End":"06:21.170","Text":"if we satisfy Rouche\u0027s theorem,"},{"Start":"06:21.170 ","End":"06:24.080","Text":"we\u0027re going to have that f_n and f will have the same number of"},{"Start":"06:24.080 ","End":"06:28.310","Text":"0s in D. But we have to show that this g,"},{"Start":"06:28.310 ","End":"06:31.270","Text":"which is g_n is less than f(z),"},{"Start":"06:31.270 ","End":"06:34.110","Text":"an absolute value on the boundary."},{"Start":"06:34.110 ","End":"06:42.470","Text":"The boundary of this disk is just Gamma r. We\u0027ll apply the absolute value here."},{"Start":"06:42.470 ","End":"06:47.420","Text":"What we\u0027ll get is the absolute value of g_n is absolute value of f_n minus"},{"Start":"06:47.420 ","End":"06:52.265","Text":"f. Now we just said that this is less than Epsilon."},{"Start":"06:52.265 ","End":"06:56.180","Text":"At least for large N it is and Epsilon was"},{"Start":"06:56.180 ","End":"07:00.785","Text":"defined to be less than f(z_min) absolute value."},{"Start":"07:00.785 ","End":"07:02.975","Text":"Actually, we could have taken equals here."},{"Start":"07:02.975 ","End":"07:08.575","Text":"Never mind. This is less than or equal to f(z) by definition of minimum."},{"Start":"07:08.575 ","End":"07:12.840","Text":"What we\u0027ve shown is that g_n is less than or equal to"},{"Start":"07:12.840 ","End":"07:17.610","Text":"f on Gamma r. That\u0027s what we have to show for Rouche\u0027s theorem."},{"Start":"07:17.610 ","End":"07:19.950","Text":"By the conclusion of Rouche\u0027s theorem,"},{"Start":"07:19.950 ","End":"07:23.930","Text":"f_n has the same number of 0s as f in this disc,"},{"Start":"07:23.930 ","End":"07:26.210","Text":"and for large N. However,"},{"Start":"07:26.210 ","End":"07:29.150","Text":"we know how many 0s f has here."},{"Start":"07:29.150 ","End":"07:31.130","Text":"It has m of them."},{"Start":"07:31.130 ","End":"07:34.280","Text":"Well, there\u0027s only 1 different 1 and that\u0027s z_0,"},{"Start":"07:34.280 ","End":"07:37.625","Text":"but it has multiplicity m and we count with multiplicity."},{"Start":"07:37.625 ","End":"07:39.109","Text":"There are m zeros,"},{"Start":"07:39.109 ","End":"07:44.210","Text":"which means that f_n also has m 0s in this disk"},{"Start":"07:44.210 ","End":"07:49.955","Text":"for r less than Rho and for big N. This is what we had to show."},{"Start":"07:49.955 ","End":"07:54.780","Text":"That\u0027s what the question asked us and so we are done."}],"ID":23778},{"Watched":false,"Name":"Exercise 1","Duration":"5m 7s","ChapterTopicVideoID":22928,"CourseChapterTopicPlaylistID":102293,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.110","Text":"In this exercise, we have to prove that there\u0027s a big N"},{"Start":"00:04.110 ","End":"00:09.585","Text":"natural number such that for all little n bigger than big N,"},{"Start":"00:09.585 ","End":"00:17.610","Text":"the equation the following equals 0 has exactly 1 solution in D,"},{"Start":"00:17.610 ","End":"00:19.665","Text":"and I better illustrate that."},{"Start":"00:19.665 ","End":"00:24.075","Text":"It\u0027s a disk centered at Pi with radius 1."},{"Start":"00:24.075 ","End":"00:31.455","Text":"Given a hint to use Hurwitz\u0027s theorem and the Taylor series expansion for sine(z)."},{"Start":"00:31.455 ","End":"00:34.200","Text":"Well, we know the Taylor series for sine(z),"},{"Start":"00:34.200 ","End":"00:37.235","Text":"which is an entire function, is the following."},{"Start":"00:37.235 ","End":"00:41.675","Text":"The powers are odd and the signs alternate and the"},{"Start":"00:41.675 ","End":"00:47.074","Text":"denominator we have the factorial of this exponent and so on to infinity."},{"Start":"00:47.074 ","End":"00:50.920","Text":"In Sigma notation, we could write it like this,"},{"Start":"00:50.920 ","End":"00:54.555","Text":"the sum from 0 to infinity, well, this."},{"Start":"00:54.555 ","End":"00:58.774","Text":"Now let\u0027s let f_n be the sum,"},{"Start":"00:58.774 ","End":"01:00.650","Text":"but not the infinite sum,"},{"Start":"01:00.650 ","End":"01:05.150","Text":"but just up to n. In Sigma notation,"},{"Start":"01:05.150 ","End":"01:07.430","Text":"instead of taking k from 0 to infinity,"},{"Start":"01:07.430 ","End":"01:14.290","Text":"we take it from 0 just to n. These are actually the partial sums of the theories."},{"Start":"01:14.290 ","End":"01:18.200","Text":"The radius of convergence of this power series,"},{"Start":"01:18.200 ","End":"01:20.510","Text":"the Taylor series is infinity,"},{"Start":"01:20.510 ","End":"01:25.210","Text":"meaning it converges everywhere in the complex plane."},{"Start":"01:25.210 ","End":"01:29.930","Text":"There\u0027s a theorem about the convergence of Taylor series is"},{"Start":"01:29.930 ","End":"01:35.555","Text":"that it\u0027s uniform on compact subsets of the domain."},{"Start":"01:35.555 ","End":"01:38.045","Text":"In other words, it converges compactly,"},{"Start":"01:38.045 ","End":"01:39.685","Text":"we can also say,"},{"Start":"01:39.685 ","End":"01:50.195","Text":"on the domain which is the whole plane C. It\u0027s just a reminder what converges compactly."},{"Start":"01:50.195 ","End":"01:51.695","Text":"For a series,"},{"Start":"01:51.695 ","End":"01:54.215","Text":"it\u0027s pretty much the same as for a sequence."},{"Start":"01:54.215 ","End":"01:56.885","Text":"We just take the sequence of partial sums,"},{"Start":"01:56.885 ","End":"01:59.150","Text":"and if this converges compactly,"},{"Start":"01:59.150 ","End":"02:00.395","Text":"the toad is this."},{"Start":"02:00.395 ","End":"02:02.570","Text":"We do that for all convergence."},{"Start":"02:02.570 ","End":"02:03.905","Text":"When we have a series,"},{"Start":"02:03.905 ","End":"02:09.094","Text":"we take the sequence of partial sums and use the definition of attribute of a sequence."},{"Start":"02:09.094 ","End":"02:16.190","Text":"In our case, the partial sums are f_n and they converge to sine(z) compactly,"},{"Start":"02:16.190 ","End":"02:18.950","Text":"in other words, uniformly in compact sets."},{"Start":"02:18.950 ","End":"02:24.980","Text":"In particular on compact subsets of D, our domain."},{"Start":"02:24.980 ","End":"02:27.020","Text":"Let us see what about the zeros?"},{"Start":"02:27.020 ","End":"02:29.165","Text":"The zeros of sine(z),"},{"Start":"02:29.165 ","End":"02:33.230","Text":"are whole multiples of Pi,"},{"Start":"02:33.230 ","End":"02:36.155","Text":"0 Pi 2Pi, 3Pi,"},{"Start":"02:36.155 ","End":"02:40.820","Text":"and the only one that\u0027s in d is z equals Pi."},{"Start":"02:40.820 ","End":"02:42.320","Text":"In other words, this is a 0,"},{"Start":"02:42.320 ","End":"02:45.365","Text":"but there\u0027s no others in this circle."},{"Start":"02:45.365 ","End":"02:52.940","Text":"Now it\u0027s a simple 0 of Order 1 or Multiplicity 1 because the derivative is not 0,"},{"Start":"02:52.940 ","End":"02:57.140","Text":"the derivative is cosine and cosine Pi it\u0027s minus 1."},{"Start":"02:57.140 ","End":"03:01.490","Text":"Anyway, it\u0027s not 0. The order of the 0 is 1."},{"Start":"03:01.490 ","End":"03:07.730","Text":"sine(z) has m equals 1 zeros in D. Next,"},{"Start":"03:07.730 ","End":"03:10.795","Text":"we\u0027re going to apply Hurwitz\u0027s theorem."},{"Start":"03:10.795 ","End":"03:13.280","Text":"I better show you the theorem."},{"Start":"03:13.280 ","End":"03:15.605","Text":"Here it is. You can pause and read it."},{"Start":"03:15.605 ","End":"03:18.380","Text":"But basically what it is is that we have,"},{"Start":"03:18.380 ","End":"03:19.895","Text":"just like in the question,"},{"Start":"03:19.895 ","End":"03:28.850","Text":"a sequence of f_n which converge to some f. We also have"},{"Start":"03:28.850 ","End":"03:32.510","Text":"this circle of radius Rho where there\u0027s"},{"Start":"03:32.510 ","End":"03:38.420","Text":"no zeros except possibly at z naught which has order m there."},{"Start":"03:38.420 ","End":"03:40.400","Text":"We also have another circle,"},{"Start":"03:40.400 ","End":"03:42.845","Text":"radius r, and in our case,"},{"Start":"03:42.845 ","End":"03:48.860","Text":"we\u0027re going to take both Rho and little r to be 1,"},{"Start":"03:48.860 ","End":"03:51.500","Text":"and we\u0027ll take z node to be Pi,"},{"Start":"03:51.500 ","End":"03:54.540","Text":"and m will be 1."},{"Start":"03:54.540 ","End":"03:57.060","Text":"Then we can apply the theorem,"},{"Start":"03:57.060 ","End":"04:04.790","Text":"and the result says that there exists some N. Well, in the theorem,"},{"Start":"04:04.790 ","End":"04:06.859","Text":"it was N_r for each r,"},{"Start":"04:06.859 ","End":"04:08.540","Text":"but we\u0027re not varying r,"},{"Start":"04:08.540 ","End":"04:10.160","Text":"r is just going to equal Rho,"},{"Start":"04:10.160 ","End":"04:11.840","Text":"which is going to equal 1,"},{"Start":"04:11.840 ","End":"04:16.505","Text":"so we just have a single N. For all N bigger than N,"},{"Start":"04:16.505 ","End":"04:26.615","Text":"f_n has exactly 1 zero in D. f_n is just like above, equal to this."},{"Start":"04:26.615 ","End":"04:31.815","Text":"Now if this function has only 1 zero,"},{"Start":"04:31.815 ","End":"04:36.470","Text":"that means that the equation where I let this thing equals"},{"Start":"04:36.470 ","End":"04:42.275","Text":"0 has exactly 1 solution in D. Now you have to be careful here."},{"Start":"04:42.275 ","End":"04:45.710","Text":"If m wasn\u0027t 1, if it was 2,"},{"Start":"04:45.710 ","End":"04:48.290","Text":"it could have 2 zeros,"},{"Start":"04:48.290 ","End":"04:51.680","Text":"but only 1 solution because of multiplicity."},{"Start":"04:51.680 ","End":"04:54.530","Text":"But when we\u0027re down to 1 zero,"},{"Start":"04:54.530 ","End":"04:56.690","Text":"it\u0027s also 1 solution."},{"Start":"04:56.690 ","End":"05:00.725","Text":"There\u0027s no multiplicity possible if the number of zeros is 1."},{"Start":"05:00.725 ","End":"05:03.110","Text":"Just pointing that out to be precise."},{"Start":"05:03.110 ","End":"05:07.910","Text":"Anyway, this is exactly what we had to prove and so we\u0027re done."}],"ID":23779},{"Watched":false,"Name":"Exercise 2","Duration":"3m 45s","ChapterTopicVideoID":22929,"CourseChapterTopicPlaylistID":102293,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/22929.jpeg","UploadDate":"2020-08-19T11:21:44.6800000","DurationForVideoObject":"PT3M45S","Description":null,"MetaTitle":"Exercise 2 - Hurwitz s Theorem: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Hurwitz s Theorem practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/complex-analysis/the-argument-principle/hurwitz-s-theorem/vid23780","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"In this exercise, we have a positive real number R for radius,"},{"Start":"00:04.980 ","End":"00:10.080","Text":"and we have to prove that there is some natural number N, such that,"},{"Start":"00:10.080 ","End":"00:12.060","Text":"for all little n bigger than it,"},{"Start":"00:12.060 ","End":"00:14.340","Text":"the equation, this one,"},{"Start":"00:14.340 ","End":"00:17.460","Text":"has no solutions in the disk D,"},{"Start":"00:17.460 ","End":"00:23.850","Text":"which is the disk centered at 0 with radius R. You can also write it this way."},{"Start":"00:23.850 ","End":"00:26.445","Text":"Here\u0027s a picture of it."},{"Start":"00:26.445 ","End":"00:30.925","Text":"Given a hint, that we have to use Hurwitz\u0027s theorem"},{"Start":"00:30.925 ","End":"00:35.360","Text":"and the Taylor series for the exponential function."},{"Start":"00:35.360 ","End":"00:38.240","Text":"Well, the exponential function is an entire function and it\u0027s"},{"Start":"00:38.240 ","End":"00:43.185","Text":"a well known Taylor series, is this."},{"Start":"00:43.185 ","End":"00:45.620","Text":"If you like it in Sigma notation,"},{"Start":"00:45.620 ","End":"00:46.700","Text":"this is how we write it."},{"Start":"00:46.700 ","End":"00:49.655","Text":"Note that we go from 0 to infinity."},{"Start":"00:49.655 ","End":"00:52.710","Text":"I\u0027m going to let f(n),"},{"Start":"00:52.710 ","End":"00:55.950","Text":"for each n be the partial sums."},{"Start":"00:55.950 ","End":"00:59.160","Text":"Instead of k going from 0 to infinity,"},{"Start":"00:59.160 ","End":"01:02.865","Text":"goes from 0 to n. These are the partial sums."},{"Start":"01:02.865 ","End":"01:07.065","Text":"They converge to e^z."},{"Start":"01:07.065 ","End":"01:11.675","Text":"Anyway, the radius of convergence is infinity because"},{"Start":"01:11.675 ","End":"01:17.035","Text":"this is the Taylor series for the entire function, e^z converges everywhere."},{"Start":"01:17.035 ","End":"01:22.235","Text":"There\u0027s a theorem that wherever a Taylor series converges,"},{"Start":"01:22.235 ","End":"01:27.559","Text":"that convergence is uniform on compact subsets of the domain."},{"Start":"01:27.559 ","End":"01:31.175","Text":"In other words, the series converges compactly."},{"Start":"01:31.175 ","End":"01:34.700","Text":"Now what that means for a series to converge compactly,"},{"Start":"01:34.700 ","End":"01:40.070","Text":"it means that the sequence of partial sums, it converges compactly."},{"Start":"01:40.070 ","End":"01:43.310","Text":"We use the definition of a sequence for the definition"},{"Start":"01:43.310 ","End":"01:46.670","Text":"of a series as far as compact convergence."},{"Start":"01:46.670 ","End":"01:51.595","Text":"So f_n, the partial sums converge to e^z,"},{"Start":"01:51.595 ","End":"01:55.095","Text":"uniformly on compact sets, and in particular,"},{"Start":"01:55.095 ","End":"01:58.860","Text":"on compact subsets of D. It converges compactly,"},{"Start":"01:58.860 ","End":"01:59.990","Text":"not just in the whole plane,"},{"Start":"01:59.990 ","End":"02:02.965","Text":"but also specifically on D. Now,"},{"Start":"02:02.965 ","End":"02:07.650","Text":"e^z has no 0\u0027s anywhere in the complex plane, and in particular,"},{"Start":"02:07.650 ","End":"02:09.390","Text":"it has no 0\u0027s in D,"},{"Start":"02:09.390 ","End":"02:12.660","Text":"so the number of 0\u0027s m is 0."},{"Start":"02:12.660 ","End":"02:15.360","Text":"It has 0, 0, so no 0\u0027s."},{"Start":"02:15.360 ","End":"02:20.910","Text":"Now, we\u0027re going to apply Hurwitz\u0027s theorem and show you what that is."},{"Start":"02:20.910 ","End":"02:27.830","Text":"It\u0027s just a reminder where we have a sequence f_n converging compactly to a"},{"Start":"02:27.830 ","End":"02:35.780","Text":"function f. We also have a circle with center at z naught and radius Rho,"},{"Start":"02:35.780 ","End":"02:38.165","Text":"that\u0027s going to be in our case,"},{"Start":"02:38.165 ","End":"02:45.680","Text":"0 and big R. The little r is also going to be big R. It could be less than or equal to,"},{"Start":"02:45.680 ","End":"02:47.240","Text":"will take it as equal to."},{"Start":"02:47.240 ","End":"02:51.005","Text":"The multiplicity in our case m, is going to be 0."},{"Start":"02:51.005 ","End":"02:54.285","Text":"Now back to our exercise."},{"Start":"02:54.285 ","End":"02:58.010","Text":"This is where I said we take the center of the circle to be 0."},{"Start":"02:58.010 ","End":"03:01.390","Text":"R and Rho are both going to be big R,"},{"Start":"03:01.390 ","End":"03:04.445","Text":"m is 0 for the multiplicity."},{"Start":"03:04.445 ","End":"03:07.280","Text":"From the conclusion of Hurwitz\u0027s theorem,"},{"Start":"03:07.280 ","End":"03:12.000","Text":"we get that there exists big N. Normally it depends on little r,"},{"Start":"03:12.000 ","End":"03:14.250","Text":"but here r is not varying,"},{"Start":"03:14.250 ","End":"03:17.405","Text":"so it\u0027s just n, such that for all little n bigger than it,"},{"Start":"03:17.405 ","End":"03:19.730","Text":"f_n has exactly 0,"},{"Start":"03:19.730 ","End":"03:22.330","Text":"0\u0027s, meaning no 0\u0027s in d,"},{"Start":"03:22.330 ","End":"03:26.075","Text":"where f_n is this partial sum."},{"Start":"03:26.075 ","End":"03:29.135","Text":"If this doesn\u0027t have any 0\u0027s,"},{"Start":"03:29.135 ","End":"03:31.850","Text":"then this doesn\u0027t have any solution."},{"Start":"03:31.850 ","End":"03:36.860","Text":"Because the solution of this would be a 0 of this."},{"Start":"03:36.860 ","End":"03:39.598","Text":"That\u0027s what it means when we take something equals 0."},{"Start":"03:39.598 ","End":"03:45.870","Text":"That\u0027s what we had to show and we\u0027ve shown it."}],"ID":23780}],"Thumbnail":null,"ID":102293},{"Name":"Existence of Logarithms and Roots of Functions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Analytic Branch of the Log of a Function","Duration":"7m 33s","ChapterTopicVideoID":22930,"CourseChapterTopicPlaylistID":102294,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.180","Text":"Here we\u0027re actually continuing a topic that we"},{"Start":"00:03.180 ","End":"00:06.480","Text":"started at another chapter on the complex integration."},{"Start":"00:06.480 ","End":"00:09.164","Text":"There was a section called primitive functions,"},{"Start":"00:09.164 ","End":"00:11.340","Text":"and we introduced the concept there."},{"Start":"00:11.340 ","End":"00:13.050","Text":"You might want to take a look at that."},{"Start":"00:13.050 ","End":"00:14.745","Text":"But really everything that we need,"},{"Start":"00:14.745 ","End":"00:17.080","Text":"I\u0027ll start afresh here."},{"Start":"00:17.080 ","End":"00:22.040","Text":"Let\u0027s get to the definition of the main concept,"},{"Start":"00:22.040 ","End":"00:27.355","Text":"f is analytic in some domain, open and connected."},{"Start":"00:27.355 ","End":"00:31.430","Text":"Then an analytic function g on the same domain"},{"Start":"00:31.430 ","End":"00:36.065","Text":"is called a branch of the logarithm of f on D,"},{"Start":"00:36.065 ","End":"00:42.085","Text":"if e^g(z) = f(z)"},{"Start":"00:42.085 ","End":"00:47.590","Text":"for all z in D. This is just a regular exponential function e to the power of."},{"Start":"00:47.590 ","End":"00:50.895","Text":"To say that g is the log of f,"},{"Start":"00:50.895 ","End":"00:56.525","Text":"means that e^g is f. This is all different from the real case."},{"Start":"00:56.525 ","End":"00:58.475","Text":"In the real case, we would just say,"},{"Start":"00:58.475 ","End":"01:01.595","Text":"okay, we\u0027ll require f to be positive,"},{"Start":"01:01.595 ","End":"01:08.420","Text":"then any positive number has a logarithm and it\u0027s well-defined."},{"Start":"01:08.420 ","End":"01:11.959","Text":"It\u0027s not multi-valued like it is in the complex case."},{"Start":"01:11.959 ","End":"01:13.310","Text":"Yeah, in the complex case,"},{"Start":"01:13.310 ","End":"01:15.868","Text":"everything is different because of this multi-valued,"},{"Start":"01:15.868 ","End":"01:18.155","Text":"and the concepts of branches."},{"Start":"01:18.155 ","End":"01:20.345","Text":"Some more terminology."},{"Start":"01:20.345 ","End":"01:24.215","Text":"Another way of saying that g is a branch of the logarithm of f,"},{"Start":"01:24.215 ","End":"01:28.115","Text":"is to say that g is an analytic logarithm of f"},{"Start":"01:28.115 ","End":"01:32.630","Text":"or under the branch of the logarithm of f. Sometimes you just get lazy,"},{"Start":"01:32.630 ","End":"01:34.265","Text":"and drop the words branch."},{"Start":"01:34.265 ","End":"01:37.760","Text":"Sometimes say g is an analytic log of f,"},{"Start":"01:37.760 ","End":"01:45.425","Text":"or g is an analytic branch of log f on D. Now we write log f. It\u0027s a bit misleading."},{"Start":"01:45.425 ","End":"01:49.250","Text":"It\u0027s not like there\u0027s a function f and then there\u0027s a branch of"},{"Start":"01:49.250 ","End":"01:53.570","Text":"logarithm and we take the logarithm of f, no it\u0027s inseparable."},{"Start":"01:53.570 ","End":"01:55.565","Text":"Log f is one concept."},{"Start":"01:55.565 ","End":"01:58.430","Text":"It\u0027s that g such that e^g is f,"},{"Start":"01:58.430 ","End":"02:02.210","Text":"it may not be a branch of logarithm such that log"},{"Start":"02:02.210 ","End":"02:06.710","Text":"f is log of f(g) in other words that might not be so."},{"Start":"02:06.710 ","End":"02:09.170","Text":"We\u0027ll see, a remark,"},{"Start":"02:09.170 ","End":"02:11.630","Text":"if g is an analytic branch of log f,"},{"Start":"02:11.630 ","End":"02:18.230","Text":"then so is g plus 2Pi i or any whole multiple of 2Pi i."},{"Start":"02:18.230 ","End":"02:23.250","Text":"That\u0027s because e^2Pi i is 1."},{"Start":"02:23.250 ","End":"02:26.899","Text":"The branch of a logarithm of a function is not unique."},{"Start":"02:26.899 ","End":"02:28.070","Text":"Once you have one branch,"},{"Start":"02:28.070 ","End":"02:32.285","Text":"you can have infinitely many more by just playing with k here."},{"Start":"02:32.285 ","End":"02:36.530","Text":"Next we come to a theorem which gives us an if and only if"},{"Start":"02:36.530 ","End":"02:40.340","Text":"condition on the existence of an analytic log."},{"Start":"02:40.340 ","End":"02:47.165","Text":"Let\u0027s say that f is analytic on a domain D. We always assume that f is not 0."},{"Start":"02:47.165 ","End":"02:49.110","Text":"I don\u0027t mean that f is not identically 0,"},{"Start":"02:49.110 ","End":"02:51.530","Text":"I mean, f is nowhere 0,"},{"Start":"02:51.530 ","End":"02:56.780","Text":"then there exists an analytic branch of logarithm of f on D,"},{"Start":"02:56.780 ","End":"03:02.010","Text":"if and only if the integral of f\u0027 over"},{"Start":"03:02.010 ","End":"03:07.380","Text":"f is 0 for each simple closed contour Gamma in D. This,"},{"Start":"03:07.380 ","End":"03:09.930","Text":"of course, exists because f is not 0;"},{"Start":"03:09.930 ","End":"03:15.335","Text":"so 1 over f is analytic and so is f\u0027 and we have an analytic function here."},{"Start":"03:15.335 ","End":"03:18.365","Text":"If this is always 0, this integral in the loop,"},{"Start":"03:18.365 ","End":"03:21.800","Text":"then that\u0027s an if and only if condition for the existence of"},{"Start":"03:21.800 ","End":"03:25.790","Text":"an analytic log of f. Now this time in this chapter,"},{"Start":"03:25.790 ","End":"03:27.890","Text":"I won\u0027t prove the theorem."},{"Start":"03:27.890 ","End":"03:34.850","Text":"But, if you go back to the chapter Complex Integration section Primitive Functions,"},{"Start":"03:34.850 ","End":"03:39.085","Text":"you\u0027ll find that there\u0027s proof there if you need it."},{"Start":"03:39.085 ","End":"03:41.530","Text":"After the theorem there\u0027s the corollary,"},{"Start":"03:41.530 ","End":"03:45.580","Text":"something derived from the theorem that is useful,"},{"Start":"03:45.580 ","End":"03:50.260","Text":"this time it\u0027s a one-way condition for the existence of a branch of"},{"Start":"03:50.260 ","End":"03:55.030","Text":"the logarithm of f if the domain happens to be simply connected."},{"Start":"03:55.030 ","End":"03:58.540","Text":"Again that f is never 0 and it\u0027s analytic,"},{"Start":"03:58.540 ","End":"04:01.945","Text":"then it has an analytic branch of the log."},{"Start":"04:01.945 ","End":"04:04.390","Text":"Next, we\u0027ll move on to an example."},{"Start":"04:04.390 ","End":"04:10.240","Text":"We have to show that there exists an analytic branch of the logarithm of this function,"},{"Start":"04:10.240 ","End":"04:15.520","Text":"f(z) equals z minus 1 over z plus 1 on the domain D,"},{"Start":"04:15.520 ","End":"04:21.695","Text":"which is all the complex numbers except for the interval from minus 1 to 1."},{"Start":"04:21.695 ","End":"04:25.550","Text":"In the diagram this is the part that\u0027s being removed,"},{"Start":"04:25.550 ","End":"04:30.500","Text":"and we\u0027ll be using the theorem we just mentioned in order to show this."},{"Start":"04:30.500 ","End":"04:36.200","Text":"Now, f is analytic and non-zero in D. The only place"},{"Start":"04:36.200 ","End":"04:45.055","Text":"the denominator can be 0 is minus 1 and that\u0027s not in D. The numerator can only be 0 at,"},{"Start":"04:45.055 ","End":"04:48.670","Text":"z=1, but 1 and minus 1 are both not included in"},{"Start":"04:48.670 ","End":"04:53.720","Text":"D. What we have to show is that this integral of f\u0027"},{"Start":"04:53.720 ","End":"05:01.280","Text":"over f is 0 for every simple closed contour Gamma that\u0027s inside D. I"},{"Start":"05:01.280 ","End":"05:05.970","Text":"claim that the points minus 1 and 1 are either"},{"Start":"05:05.970 ","End":"05:11.015","Text":"both inside or both outside the contour Gamma."},{"Start":"05:11.015 ","End":"05:14.885","Text":"In other words, not one inside and one outside."},{"Start":"05:14.885 ","End":"05:18.860","Text":"The reason for that is if you did have one inside and one outside,"},{"Start":"05:18.860 ","End":"05:24.680","Text":"then the curve Gamma would intersect this interval from minus 1 to 1."},{"Start":"05:24.680 ","End":"05:30.455","Text":"If any path joining an inside point to an outside point has to cross the curve."},{"Start":"05:30.455 ","End":"05:33.830","Text":"That will contradict the fact that Gamma is in D,"},{"Start":"05:33.830 ","End":"05:39.290","Text":"and D does not include the integral from minus 1 to 1,"},{"Start":"05:39.290 ","End":"05:41.390","Text":"so Gamma can\u0027t cut it."},{"Start":"05:41.390 ","End":"05:46.665","Text":"As far as zeros and poles of f go,"},{"Start":"05:46.665 ","End":"05:49.555","Text":"it\u0027s clear that there is a simple pole at"},{"Start":"05:49.555 ","End":"05:54.010","Text":"z=minus 1 and a simple 0 at z=1, and that\u0027s it."},{"Start":"05:54.010 ","End":"05:56.695","Text":"Those are the poles and zeros."},{"Start":"05:56.695 ","End":"06:00.280","Text":"Let\u0027s let N and P be the number of zeros and poles"},{"Start":"06:00.280 ","End":"06:06.080","Text":"respectively of f that are inside the given curve Gamma,"},{"Start":"06:06.080 ","End":"06:07.965","Text":"and it will vary for Gamma."},{"Start":"06:07.965 ","End":"06:10.090","Text":"By the argument principle,"},{"Start":"06:10.090 ","End":"06:13.945","Text":"we have this formula that the integral of f\u0027 over f is 2Pi i"},{"Start":"06:13.945 ","End":"06:18.875","Text":"times N minus P. Let\u0027s distinguish the 2 cases for Gamma."},{"Start":"06:18.875 ","End":"06:23.124","Text":"If the points minus 1 and 1 are both outside Gamma,"},{"Start":"06:23.124 ","End":"06:25.895","Text":"then here is a diagram."},{"Start":"06:25.895 ","End":"06:31.475","Text":"The points minus 1 and 1 are not inside the curve, not enclosed by."},{"Start":"06:31.475 ","End":"06:37.715","Text":"In that case, there\u0027s no zeros or poles inside to N=0, P=0."},{"Start":"06:37.715 ","End":"06:39.990","Text":"Then the integral which is"},{"Start":"06:39.990 ","End":"06:46.609","Text":"2Pi i N minus P is just going to be 0 because N and P are both 0."},{"Start":"06:46.609 ","End":"06:48.485","Text":"Now in the other case,"},{"Start":"06:48.485 ","End":"06:53.890","Text":"where these are both inside Gamma, again the diagram."},{"Start":"06:53.890 ","End":"06:56.930","Text":"In that case we have that this integral,"},{"Start":"06:56.930 ","End":"06:59.645","Text":"which is again 2Pi i N minus P,"},{"Start":"06:59.645 ","End":"07:03.905","Text":"but this time N=1 and P=1,"},{"Start":"07:03.905 ","End":"07:06.110","Text":"because one zero, one pole."},{"Start":"07:06.110 ","End":"07:10.975","Text":"This computation now comes out to be 1 minus 1."},{"Start":"07:10.975 ","End":"07:12.920","Text":"It\u0027s a different computation,"},{"Start":"07:12.920 ","End":"07:14.975","Text":"but it still gives us 0."},{"Start":"07:14.975 ","End":"07:18.200","Text":"In both cases, we get that the integral is"},{"Start":"07:18.200 ","End":"07:24.110","Text":"0 and since this integral is 0 for all curves Gamma,"},{"Start":"07:24.110 ","End":"07:30.155","Text":"that means by the theorem that f does have an analytic log in D,"},{"Start":"07:30.155 ","End":"07:33.570","Text":"and that\u0027s what we had to show and we\u0027re done."}],"ID":23781},{"Watched":false,"Name":"Analytic n-th root of a Function","Duration":"4m 22s","ChapterTopicVideoID":22931,"CourseChapterTopicPlaylistID":102294,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.400","Text":"In this clip, we\u0027ll talk about the concept of"},{"Start":"00:03.400 ","End":"00:07.520","Text":"an analytic N-th root of an analytic function."},{"Start":"00:07.520 ","End":"00:11.770","Text":"Just like in the case of the analytic logarithm,"},{"Start":"00:11.770 ","End":"00:16.960","Text":"it\u0027s not so simple in the complex cases in the real case because an N-th root,"},{"Start":"00:16.960 ","End":"00:23.635","Text":"like a square root or a cube root could be multi-valued and we have to choose a branch."},{"Start":"00:23.635 ","End":"00:25.195","Text":"I\u0027ll start with the definition."},{"Start":"00:25.195 ","End":"00:31.965","Text":"F is an analytic function on a domain D and its open and connected set,"},{"Start":"00:31.965 ","End":"00:34.060","Text":"subset of C and then n be"},{"Start":"00:34.060 ","End":"00:38.875","Text":"a positive integer though typically will take n to be bigger or equal to 2."},{"Start":"00:38.875 ","End":"00:42.530","Text":"We don\u0027t take the 1-th of first route usually."},{"Start":"00:42.530 ","End":"00:50.320","Text":"Then an analytic function g also on D is called an analytic n-th root of f on D,"},{"Start":"00:50.320 ","End":"00:58.490","Text":"if g^n gives us our function f in D. When n is 2 or 3,"},{"Start":"00:58.490 ","End":"01:04.505","Text":"we say square root and cube root and then fourth or fifth or sixth root and so on."},{"Start":"01:04.505 ","End":"01:09.290","Text":"We write it as g(z) equals the symbol of"},{"Start":"01:09.290 ","End":"01:14.570","Text":"nth root of f(z) on D. If we have 1 analytic n-th root,"},{"Start":"01:14.570 ","End":"01:19.340","Text":"we can also multiply it by a root of unity."},{"Start":"01:19.340 ","End":"01:24.180","Text":"This would be an n-th root of 1 and then"},{"Start":"01:24.180 ","End":"01:29.975","Text":"this thing will also be an n-th root of f because if you raise this to the power of n,"},{"Start":"01:29.975 ","End":"01:32.420","Text":"this to the power of n is 1."},{"Start":"01:32.420 ","End":"01:37.700","Text":"We can actually get up to n different ones if we take here k to be 0,"},{"Start":"01:37.700 ","End":"01:39.365","Text":"1 up to n-1,"},{"Start":"01:39.365 ","End":"01:42.395","Text":"and then it starts looping around on itself."},{"Start":"01:42.395 ","End":"01:47.510","Text":"There\u0027ll be 2 square roots and 3 cube roots and so on."},{"Start":"01:47.510 ","End":"01:49.280","Text":"Now as an example,"},{"Start":"01:49.280 ","End":"01:50.389","Text":"it\u0027s a bit artificial,"},{"Start":"01:50.389 ","End":"01:51.935","Text":"but just to give you an idea,"},{"Start":"01:51.935 ","End":"01:54.770","Text":"if we take f(z) to be z^2,"},{"Start":"01:54.770 ","End":"01:58.640","Text":"then it has an analytic square root, second root,"},{"Start":"01:58.640 ","End":"02:05.255","Text":"which is g(z)=z because g(z)^2 gives us f(z)."},{"Start":"02:05.255 ","End":"02:06.875","Text":"But like I said, it\u0027s not unique."},{"Start":"02:06.875 ","End":"02:10.820","Text":"We could have also taken g(z) = -z and that\u0027s also"},{"Start":"02:10.820 ","End":"02:15.395","Text":"a square root because -z^2 is also z^2."},{"Start":"02:15.395 ","End":"02:19.420","Text":"Another example, f(z) = sine ^4 z."},{"Start":"02:19.420 ","End":"02:21.500","Text":"It has an analytic fourth root,"},{"Start":"02:21.500 ","End":"02:27.080","Text":"just sine z because sine z^4 is sine 4(z)."},{"Start":"02:27.080 ","End":"02:29.900","Text":"Again, it\u0027s not unique."},{"Start":"02:29.900 ","End":"02:36.500","Text":"We could take g(z) = iz or -z or -iz."},{"Start":"02:36.500 ","End":"02:41.695","Text":"Also ^4, it will give us just sine^4z, which is f(z)."},{"Start":"02:41.695 ","End":"02:43.325","Text":"Now, with theorem,"},{"Start":"02:43.325 ","End":"02:47.795","Text":"if we know that f has a non-analytic logarithm on d,"},{"Start":"02:47.795 ","End":"02:50.960","Text":"then f has an analytic n-th root on D for"},{"Start":"02:50.960 ","End":"02:55.550","Text":"all positive integers n. It\u0027s not a difficult proof,"},{"Start":"02:55.550 ","End":"02:56.765","Text":"so we\u0027ll do it here."},{"Start":"02:56.765 ","End":"03:04.280","Text":"We need to find some analytic g such that g^n is f. We\u0027re told that it has"},{"Start":"03:04.280 ","End":"03:09.280","Text":"an analytic logarithm on D. Let\u0027s call that log f. That\u0027s"},{"Start":"03:09.280 ","End":"03:11.600","Text":"the function which is the analytic log of f which"},{"Start":"03:11.600 ","End":"03:14.690","Text":"exists by the definition of analytic log,"},{"Start":"03:14.690 ","End":"03:21.640","Text":"we get that e^log f(z) gives us f(z)."},{"Start":"03:21.640 ","End":"03:24.705","Text":"Now I\u0027m going to define an n-th root of f,"},{"Start":"03:24.705 ","End":"03:32.540","Text":"defined g(z) to be e^1 times log f. This is the g we\u0027re looking for."},{"Start":"03:32.540 ","End":"03:38.450","Text":"In other words, I have to show that g^n is f. I didn\u0027t write it but note"},{"Start":"03:38.450 ","End":"03:45.075","Text":"that this is also analytic because ln(f) is analytic."},{"Start":"03:45.075 ","End":"03:48.870","Text":"Divide it by the constant n. It\u0027s still analytic."},{"Start":"03:48.870 ","End":"03:54.500","Text":"Each of the power of is analytic because composition of analytic functions is analytic."},{"Start":"03:54.500 ","End":"04:00.080","Text":"Let\u0027s just check that this g^n is = f. G is equal to this,"},{"Start":"04:00.080 ","End":"04:07.130","Text":"raise it to the power of n. We can multiply the exponents n times 1 times this."},{"Start":"04:07.130 ","End":"04:11.545","Text":"The n cancels, and it\u0027s just e^log f(z)."},{"Start":"04:11.545 ","End":"04:14.675","Text":"By definition, here it is."},{"Start":"04:14.675 ","End":"04:16.235","Text":"This is = f(z),"},{"Start":"04:16.235 ","End":"04:18.755","Text":"so g^n is f,"},{"Start":"04:18.755 ","End":"04:23.490","Text":"and that\u0027s what we had to find. We\u0027re done."}],"ID":23782},{"Watched":false,"Name":"Exercise 1","Duration":"2m 12s","ChapterTopicVideoID":22932,"CourseChapterTopicPlaylistID":102294,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.775","Text":"In this exercise, we have the function f(z) equals z minus 1 over z plus 1."},{"Start":"00:05.775 ","End":"00:07.950","Text":"We have to show that this function has"},{"Start":"00:07.950 ","End":"00:11.745","Text":"an analytic branch of the logarithm in the domain D,"},{"Start":"00:11.745 ","End":"00:13.905","Text":"which is the complex plane."},{"Start":"00:13.905 ","End":"00:18.330","Text":"But, we cut out the interval from minus 1 down"},{"Start":"00:18.330 ","End":"00:23.085","Text":"to minus infinity and the interval from 1 to infinity, those are removed."},{"Start":"00:23.085 ","End":"00:26.355","Text":"We have an open connected set and that\u0027s D."},{"Start":"00:26.355 ","End":"00:30.360","Text":"We\u0027re going to be using the main theorem for this and the theorem to remind you"},{"Start":"00:30.360 ","End":"00:35.220","Text":"says that an analytic function which is not 0 anywhere on D has"},{"Start":"00:35.220 ","End":"00:40.850","Text":"an analytic log if and only if for each simple closed contour Gamma,"},{"Start":"00:40.850 ","End":"00:42.860","Text":"in D this integral is 0,"},{"Start":"00:42.860 ","End":"00:46.378","Text":"the integral of f\u0027 over f. Now the zeros,"},{"Start":"00:46.378 ","End":"00:53.000","Text":"and poles of f. It has a 0 at z equals 1 and it has a pole at"},{"Start":"00:53.000 ","End":"01:00.050","Text":"z equals minus 1 and neither of these is in Gamma, or its interior."},{"Start":"01:00.050 ","End":"01:06.170","Text":"Because if we have a loop somewhere and it has 1 in its interior,"},{"Start":"01:06.170 ","End":"01:08.560","Text":"it has to cut this ray somewhere."},{"Start":"01:08.560 ","End":"01:11.540","Text":"They\u0027re allowed to do that similarly with minus 1."},{"Start":"01:11.540 ","End":"01:14.120","Text":"Basically, we have something like this."},{"Start":"01:14.120 ","End":"01:17.840","Text":"It can\u0027t loop around minus 1 or 1."},{"Start":"01:17.840 ","End":"01:22.070","Text":"What this implies is that f is analytic,"},{"Start":"01:22.070 ","End":"01:24.818","Text":"and non-zero in Gamma,"},{"Start":"01:24.818 ","End":"01:29.115","Text":"and its interior because inside here we don\u0027t have 1 or minus 1."},{"Start":"01:29.115 ","End":"01:31.550","Text":"Minus 1 is the only place where it\u0027s not analytic."},{"Start":"01:31.550 ","End":"01:37.885","Text":"That\u0027s the pole and also it\u0027s non-zero because only at 1 can it be 0, so like so."},{"Start":"01:37.885 ","End":"01:42.720","Text":"Hence, f\u0027 over f is analytic in Gamma and"},{"Start":"01:42.720 ","End":"01:48.485","Text":"its interior because f\u0027 is on analytic and f is analytic and non-zero."},{"Start":"01:48.485 ","End":"01:54.260","Text":"That means that we can apply the Cauchy-Goursat theorem to the analytic function"},{"Start":"01:54.260 ","End":"02:00.740","Text":"f\u0027 over f on the contour Gamma and get that this is equal to 0."},{"Start":"02:00.740 ","End":"02:04.040","Text":"Now that\u0027s exactly what we wanted because this is for each Gamma"},{"Start":"02:04.040 ","End":"02:07.490","Text":"in the domain D. So by the theorem,"},{"Start":"02:07.490 ","End":"02:12.540","Text":"f has an analytic logarithm and that\u0027s the end of the exercise."}],"ID":23783},{"Watched":false,"Name":"Exercise 2","Duration":"3m 37s","ChapterTopicVideoID":22933,"CourseChapterTopicPlaylistID":102294,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"In this exercise, we have the domain D,"},{"Start":"00:03.390 ","End":"00:05.895","Text":"which is essentially the complex plane,"},{"Start":"00:05.895 ","End":"00:12.255","Text":"except that we cut out the interval from minus i to i inclusive,"},{"Start":"00:12.255 ","End":"00:14.265","Text":"as in the diagram."},{"Start":"00:14.265 ","End":"00:19.350","Text":"Part 1, we have to prove that the function z minus i over z plus"},{"Start":"00:19.350 ","End":"00:24.480","Text":"i has an analytic branch of the logarithm on the domain D."},{"Start":"00:24.480 ","End":"00:31.050","Text":"Secondly the function z^2 times the other function has"},{"Start":"00:31.050 ","End":"00:38.584","Text":"an analytic square root on the same D. We\u0027ll use the main theorem for this section."},{"Start":"00:38.584 ","End":"00:41.030","Text":"I won\u0027t reread it, I\u0027ll just put it here."},{"Start":"00:41.030 ","End":"00:48.080","Text":"We have to show that this integral is 0 for all simple closed contours,"},{"Start":"00:48.080 ","End":"00:49.655","Text":"and then we\u0027ll get our result."},{"Start":"00:49.655 ","End":"00:59.600","Text":"Now note that f has a 0 at i and a pole minus i and those are the only zeros and poles in"},{"Start":"00:59.600 ","End":"01:03.020","Text":"the whole plane and neither of these is in D because we\u0027ve cut"},{"Start":"01:03.020 ","End":"01:06.970","Text":"them out of D. That means that f is"},{"Start":"01:06.970 ","End":"01:15.790","Text":"non-zero and analytic in D. I mean we\u0027ve removed the only bad point which is the minus i."},{"Start":"01:15.790 ","End":"01:21.275","Text":"Let Gamma be a simple closed contour in D. We have to show that this integral is 0,"},{"Start":"01:21.275 ","End":"01:26.235","Text":"i and minus i are both inside Gamma or both outside Gamma."},{"Start":"01:26.235 ","End":"01:28.320","Text":"That\u0027s what I\u0027m claiming."},{"Start":"01:28.320 ","End":"01:30.330","Text":"If you think about it,"},{"Start":"01:30.330 ","End":"01:33.790","Text":"if 1 of these is inside and 1 of these is outside,"},{"Start":"01:33.790 ","End":"01:37.980","Text":"then this segment will intersect the curve or the curve will cut this segment,"},{"Start":"01:37.980 ","End":"01:43.240","Text":"so they have to both be inside or both be outside."},{"Start":"01:43.240 ","End":"01:45.744","Text":"Drawn a picture of each of the cases."},{"Start":"01:45.744 ","End":"01:47.920","Text":"I mean, we can\u0027t have one in and one out."},{"Start":"01:47.920 ","End":"01:53.630","Text":"Let N and P be the number of zeros and poles respectively inside Gamma,"},{"Start":"01:53.630 ","End":"01:56.910","Text":"zeros and poles of little f, I mean."},{"Start":"01:57.800 ","End":"02:00.575","Text":"If we take this case,"},{"Start":"02:00.575 ","End":"02:02.720","Text":"then N and P are both 1."},{"Start":"02:02.720 ","End":"02:05.030","Text":"We have one 0 and one pole."},{"Start":"02:05.030 ","End":"02:08.660","Text":"In this case they are both 0 we have no zeros,"},{"Start":"02:08.660 ","End":"02:11.050","Text":"and no poles inside Gamma."},{"Start":"02:11.050 ","End":"02:12.705","Text":"But in either case,"},{"Start":"02:12.705 ","End":"02:18.480","Text":"either 1 minus 1 or 0 minus 0 gives us 0."},{"Start":"02:18.480 ","End":"02:21.110","Text":"By the argument principle,"},{"Start":"02:21.110 ","End":"02:22.700","Text":"the integral we\u0027re looking for,"},{"Start":"02:22.700 ","End":"02:29.450","Text":"integral of f\u0027 over f is 2Pi i times N minus P. In either case this is 0."},{"Start":"02:29.450 ","End":"02:33.410","Text":"By the theorem, f has an analytical logarithm,"},{"Start":"02:33.410 ","End":"02:36.115","Text":"and that concludes part 1."},{"Start":"02:36.115 ","End":"02:42.229","Text":"For part 2 we\u0027ll need the other theorem that if f has an analytic log,"},{"Start":"02:42.229 ","End":"02:44.675","Text":"then it has an analytic nth root for each"},{"Start":"02:44.675 ","End":"02:49.023","Text":"N. All we have to do is apply this theorem with n equals 2,"},{"Start":"02:49.023 ","End":"02:51.523","Text":"and that gives us that little f,"},{"Start":"02:51.523 ","End":"02:53.330","Text":"it has an analytic square root,"},{"Start":"02:53.330 ","End":"02:58.220","Text":"call that g on D. But we\u0027re talking about little f, where is it?"},{"Start":"02:58.220 ","End":"03:00.750","Text":"Yeah, we need for big F,"},{"Start":"03:00.750 ","End":"03:04.715","Text":"which is z^2 times this g is the square root of little f,"},{"Start":"03:04.715 ","End":"03:08.195","Text":"so g^2 is little f(z)."},{"Start":"03:08.195 ","End":"03:12.410","Text":"Now let h be z times g(z)."},{"Start":"03:12.410 ","End":"03:14.419","Text":"If you want you could put a minus also,"},{"Start":"03:14.419 ","End":"03:19.376","Text":"I\u0027m just putting a plus or minus for illustration."},{"Start":"03:19.376 ","End":"03:23.230","Text":"h^2 gives us exactly big F. That"},{"Start":"03:23.230 ","End":"03:27.754","Text":"means that h is an analytic square root of f. Of course h is analytic,"},{"Start":"03:27.754 ","End":"03:29.990","Text":"also, I mean, if g is analytic,"},{"Start":"03:29.990 ","End":"03:33.490","Text":"then multiplying by z will still be analytic."},{"Start":"03:33.490 ","End":"03:35.535","Text":"That concludes part 2,"},{"Start":"03:35.535 ","End":"03:37.990","Text":"and we are done."}],"ID":23784},{"Watched":false,"Name":"Exercise 3","Duration":"7m 26s","ChapterTopicVideoID":22934,"CourseChapterTopicPlaylistID":102294,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.345","Text":"In this exercise, we have the domain D,"},{"Start":"00:03.345 ","End":"00:05.654","Text":"which is the complex plane,"},{"Start":"00:05.654 ","End":"00:10.335","Text":"where we remove the interval from minus 1 to 1."},{"Start":"00:10.335 ","End":"00:12.960","Text":"This is the red part that\u0027s cut out."},{"Start":"00:12.960 ","End":"00:15.000","Text":"It\u0027s an open and connected set,"},{"Start":"00:15.000 ","End":"00:20.535","Text":"and that f(z) equal 1 minus z squared on this domain D,"},{"Start":"00:20.535 ","End":"00:24.090","Text":"it\u0027s a polynomial, so no problem defining that everywhere."},{"Start":"00:24.090 ","End":"00:25.350","Text":"We have two things to prove,"},{"Start":"00:25.350 ","End":"00:33.060","Text":"first of all that f(z) does not have an analytic logarithm on D. In part 2,"},{"Start":"00:33.060 ","End":"00:36.740","Text":"we\u0027re going to prove that f(z) does have an analytic square root"},{"Start":"00:36.740 ","End":"00:40.640","Text":"on D. There\u0027s no contradiction, or anything here."},{"Start":"00:40.640 ","End":"00:43.865","Text":"If f does have an analytic logarithm,"},{"Start":"00:43.865 ","End":"00:46.205","Text":"then it also has an analytic square roots,"},{"Start":"00:46.205 ","End":"00:47.990","Text":"an analytic cube root,"},{"Start":"00:47.990 ","End":"00:50.045","Text":"fourth root, any order route."},{"Start":"00:50.045 ","End":"00:53.435","Text":"But if f(z) doesn\u0027t have an analytic logarithm,"},{"Start":"00:53.435 ","End":"00:58.295","Text":"it may or may not have analytics square root or cube root or whatever."},{"Start":"00:58.295 ","End":"01:00.545","Text":"Doesn\u0027t contradict either way."},{"Start":"01:00.545 ","End":"01:03.560","Text":"Now we have our theorem on how to tell whether or"},{"Start":"01:03.560 ","End":"01:07.310","Text":"not the function has an analytic logarithm on a domain."},{"Start":"01:07.310 ","End":"01:14.660","Text":"That is, if and only if this integral is 0 on all contours Gamma in D,"},{"Start":"01:14.660 ","End":"01:16.040","Text":"since it\u0027s a both way thing,"},{"Start":"01:16.040 ","End":"01:18.155","Text":"we can use it to prove, or disprove."},{"Start":"01:18.155 ","End":"01:21.380","Text":"In this case, we want to say that it doesn\u0027t have an analytic log,"},{"Start":"01:21.380 ","End":"01:27.220","Text":"so we just have to bring 1 counter-example of a Gamma which doesn\u0027t satisfy this."},{"Start":"01:27.220 ","End":"01:33.705","Text":"Now, in the whole plane f has zeros only at 1 and minus 1."},{"Start":"01:33.705 ","End":"01:37.730","Text":"These are not in D because in D we\u0027ve cut them out."},{"Start":"01:37.730 ","End":"01:42.500","Text":"We\u0027ve removed this whole interval so that f is not equal to"},{"Start":"01:42.500 ","End":"01:48.030","Text":"0 anywhere in D and its analytic, analytic and non-zero."},{"Start":"01:48.030 ","End":"01:49.835","Text":"Now let us take a curve,"},{"Start":"01:49.835 ","End":"01:51.470","Text":"call it big Gamma,"},{"Start":"01:51.470 ","End":"01:54.995","Text":"which wraps around this interval."},{"Start":"01:54.995 ","End":"02:02.175","Text":"For example, we could take a circle of radius 2 around 0."},{"Start":"02:02.175 ","End":"02:04.635","Text":"That will certainly wrap around this."},{"Start":"02:04.635 ","End":"02:07.770","Text":"This will be our counterexample 1 and minus 1,"},{"Start":"02:07.770 ","End":"02:09.015","Text":"or both inside Gamma."},{"Start":"02:09.015 ","End":"02:11.470","Text":"That\u0027s how I chose this Gamma."},{"Start":"02:11.470 ","End":"02:18.740","Text":"Let\u0027s count the number of zeros and poles inside Gamma for the function f. Well,"},{"Start":"02:18.740 ","End":"02:22.775","Text":"it\u0027s clear there are two zeros and no poles, no poles anywhere."},{"Start":"02:22.775 ","End":"02:24.920","Text":"As we said, its function has 2 zeros."},{"Start":"02:24.920 ","End":"02:30.310","Text":"N is 2 P is 0 to N minus P is 2 in this Gamma."},{"Start":"02:30.310 ","End":"02:33.455","Text":"If that\u0027s the case by the argument principle,"},{"Start":"02:33.455 ","End":"02:37.040","Text":"this integral is 2Pi i(N minus P),"},{"Start":"02:37.040 ","End":"02:41.659","Text":"which comes out to be non-zero because N minus P is not 0,"},{"Start":"02:41.659 ","End":"02:43.610","Text":"comes out to be 4Pi IN this case,"},{"Start":"02:43.610 ","End":"02:45.170","Text":"in any event non-zero."},{"Start":"02:45.170 ","End":"02:47.210","Text":"So by the theorem,"},{"Start":"02:47.210 ","End":"02:49.730","Text":"f doesn\u0027t have an analytic logarithm."},{"Start":"02:49.730 ","End":"02:52.865","Text":"For part 2, we\u0027re going to use the other theorem."},{"Start":"02:52.865 ","End":"02:56.000","Text":"If f has an analytic logarithm on D,"},{"Start":"02:56.000 ","End":"03:03.330","Text":"then f has an analytic and root of D for all positive integers N. Now you might say,"},{"Start":"03:03.330 ","End":"03:06.650","Text":"in our case, f(z) equals 1 minus z squared,"},{"Start":"03:06.650 ","End":"03:12.260","Text":"and it doesn\u0027t have an analytic logarithm on D. We just showed that in part 1."},{"Start":"03:12.260 ","End":"03:16.460","Text":"The thing is we\u0027re going to use this but not for f, for something else."},{"Start":"03:16.460 ","End":"03:20.300","Text":"The trouble we had in part 1 was that for f,"},{"Start":"03:20.300 ","End":"03:28.370","Text":"the number of zeros minus the number of poles on some Gamma was not equal to 0,"},{"Start":"03:28.370 ","End":"03:29.615","Text":"was equal to 2."},{"Start":"03:29.615 ","End":"03:38.645","Text":"What we can do then is use a trick if we write f(z) as 1 minus z all squared times g(z),"},{"Start":"03:38.645 ","End":"03:43.470","Text":"where g(z) is this and check that this works out to this"},{"Start":"03:43.470 ","End":"03:48.290","Text":"because 1 minus z over 1 minus z cancels and we\u0027re left with 1 minus z,"},{"Start":"03:48.290 ","End":"03:51.830","Text":"1 plus z, which is 1 minus z squared."},{"Start":"03:51.830 ","End":"03:59.960","Text":"We write the f(z) this way and I claim that g(z) does satisfy N minus P equals 0,"},{"Start":"03:59.960 ","End":"04:04.985","Text":"not just this particular Gamma for all Gamma D. Just showing you the plan here."},{"Start":"04:04.985 ","End":"04:08.420","Text":"Assuming that we show this and we will show this,"},{"Start":"04:08.420 ","End":"04:10.745","Text":"then we\u0027ll be able to say that, yeah,"},{"Start":"04:10.745 ","End":"04:13.699","Text":"by this theorem applied to g,"},{"Start":"04:13.699 ","End":"04:18.935","Text":"the integral of g\u0027 over g is 2 Pi i (N minus P)."},{"Start":"04:18.935 ","End":"04:23.960","Text":"For this function, g N minus P is 0, so we\u0027ll get 0."},{"Start":"04:23.960 ","End":"04:27.395","Text":"We can apply the theorem to g, the first theorem,"},{"Start":"04:27.395 ","End":"04:31.340","Text":"and conclude that g has an analytic logarithm in D,"},{"Start":"04:31.340 ","End":"04:34.550","Text":"which means from the second theorem that g has"},{"Start":"04:34.550 ","End":"04:38.375","Text":"an analytic square root of analytic nth root for any N,"},{"Start":"04:38.375 ","End":"04:39.820","Text":"take N equals 2."},{"Start":"04:39.820 ","End":"04:42.455","Text":"An analytics square root g(z)."},{"Start":"04:42.455 ","End":"04:45.450","Text":"Now, define h(z) to equal"},{"Start":"04:45.450 ","End":"04:51.515","Text":"1 minus z times the analytics square root of g that we found here."},{"Start":"04:51.515 ","End":"04:55.220","Text":"Then h(z) squared is f(z),"},{"Start":"04:55.220 ","End":"05:00.265","Text":"because the square root of g(z) squared is g(z)."},{"Start":"05:00.265 ","End":"05:07.170","Text":"By definition, then we have g(z) times 1 minus z squared."},{"Start":"05:07.170 ","End":"05:11.580","Text":"From here, this is f(z)."},{"Start":"05:11.580 ","End":"05:15.315","Text":"That means that because h(z) squared is f(z),"},{"Start":"05:15.315 ","End":"05:20.390","Text":"that h(z) is the square root of f(z) or 1 of the analytics square roots,"},{"Start":"05:20.390 ","End":"05:23.120","Text":"which means that f has an another take square root."},{"Start":"05:23.120 ","End":"05:25.065","Text":"That was the plan."},{"Start":"05:25.065 ","End":"05:29.000","Text":"In fact it is, most of it was just a couple of details."},{"Start":"05:29.000 ","End":"05:33.290","Text":"Or really the only part that\u0027s missing is"},{"Start":"05:33.290 ","End":"05:37.700","Text":"this part where we have to show that N minus P is"},{"Start":"05:37.700 ","End":"05:46.580","Text":"0 for the function g on any Gamma in D. If we show that,"},{"Start":"05:46.580 ","End":"05:48.640","Text":"then we really are done."},{"Start":"05:48.640 ","End":"05:54.050","Text":"Now, recall that definition of g was this."},{"Start":"05:54.050 ","End":"05:56.540","Text":"You can even still see it up here."},{"Start":"05:56.540 ","End":"06:05.765","Text":"Yeah. If we look for zeros and poles of g and the whole plane,"},{"Start":"06:05.765 ","End":"06:13.285","Text":"then we see it\u0027s just a 0 at z equals minus 1 and a pole at z equals 1."},{"Start":"06:13.285 ","End":"06:20.420","Text":"Now both of these minus 1 on 1 and not in D. In D there are no zeros or poles."},{"Start":"06:20.420 ","End":"06:24.650","Text":"Also, g is non-zero whenever 0 in D. So that"},{"Start":"06:24.650 ","End":"06:29.465","Text":"means that if Gamma is a simple closed contour in D,"},{"Start":"06:29.465 ","End":"06:31.680","Text":"use a picture to explain,"},{"Start":"06:31.680 ","End":"06:37.550","Text":"then both of these minus 1 and 1 are either both outside or both"},{"Start":"06:37.550 ","End":"06:45.200","Text":"inside because the contour Gamma can\u0027t intersect the interval from minus 1 to 1."},{"Start":"06:45.200 ","End":"06:49.520","Text":"It either goes around it or misses it completely."},{"Start":"06:49.520 ","End":"06:54.005","Text":"But this whole segment has to be completely inside, a completely outside."},{"Start":"06:54.005 ","End":"06:56.825","Text":"I just wrote what I said here."},{"Start":"06:56.825 ","End":"06:59.780","Text":"At N and P be the number of zeros and poles"},{"Start":"06:59.780 ","End":"07:04.080","Text":"respectively of g inside Gamma, there\u0027s two cases."},{"Start":"07:04.080 ","End":"07:07.920","Text":"In this case we have N equals 1 and P equals 1."},{"Start":"07:07.920 ","End":"07:09.615","Text":"One zero one pole."},{"Start":"07:09.615 ","End":"07:11.959","Text":"In the other case, everything\u0027s outside,"},{"Start":"07:11.959 ","End":"07:14.105","Text":"so no zeros, no poles."},{"Start":"07:14.105 ","End":"07:17.810","Text":"But whether it\u0027s 1 minus 1 or 0 minus 0,"},{"Start":"07:17.810 ","End":"07:20.780","Text":"both cases, it\u0027s 0."},{"Start":"07:20.780 ","End":"07:23.990","Text":"That\u0027s the missing piece that we had to show,"},{"Start":"07:23.990 ","End":"07:26.640","Text":"and so we are done."}],"ID":23785},{"Watched":false,"Name":"Exercise 4","Duration":"6m 28s","ChapterTopicVideoID":22935,"CourseChapterTopicPlaylistID":102294,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"Here we have a domain D,"},{"Start":"00:02.460 ","End":"00:04.710","Text":"which is described as follows,"},{"Start":"00:04.710 ","End":"00:07.065","Text":"and you can see it in the diagram."},{"Start":"00:07.065 ","End":"00:11.460","Text":"It\u0027s everything outside the circle of radius 4."},{"Start":"00:11.460 ","End":"00:13.950","Text":"We have a function f defined on D,"},{"Start":"00:13.950 ","End":"00:18.435","Text":"which is (z-2)^2 minus 4. 2 parts."},{"Start":"00:18.435 ","End":"00:24.810","Text":"First of all, prove that f doesn\u0027t have an analytic logarithm on D, and secondly,"},{"Start":"00:24.810 ","End":"00:28.815","Text":"prove that it does have an analytic square root on D."},{"Start":"00:28.815 ","End":"00:33.900","Text":"They\u0027ll remind you of our main theorem is that an analytic function f,"},{"Start":"00:33.900 ","End":"00:41.815","Text":"which is nonzero 1D has an analytic branch of the logarithm on D if and only if"},{"Start":"00:41.815 ","End":"00:50.465","Text":"this integral is 0 for each simple closed contour Gamma in D. The integral of f\u0027 over f,"},{"Start":"00:50.465 ","End":"00:57.410","Text":"which is also analytic because f\u0027 is analytic and f is analytic and non-zero."},{"Start":"00:57.410 ","End":"00:59.524","Text":"So the quotient is analytic."},{"Start":"00:59.524 ","End":"01:01.610","Text":"Let\u0027s get to part 1."},{"Start":"01:01.610 ","End":"01:03.890","Text":"Now if we simplify this expression,"},{"Start":"01:03.890 ","End":"01:08.015","Text":"it just comes out to be Z times Z minus 4."},{"Start":"01:08.015 ","End":"01:16.235","Text":"Note that the only zeros of f in the whole plane are 0 and 4,"},{"Start":"01:16.235 ","End":"01:23.765","Text":"but these are outside of D. D does not include the circle itself."},{"Start":"01:23.765 ","End":"01:27.785","Text":"We have F which is non-zero and analytic,"},{"Start":"01:27.785 ","End":"01:30.050","Text":"analytic clearly it\u0027s a polynomial,"},{"Start":"01:30.050 ","End":"01:32.345","Text":"and in D, and now,"},{"Start":"01:32.345 ","End":"01:34.415","Text":"let\u0027s choose a particular Gamma,"},{"Start":"01:34.415 ","End":"01:35.570","Text":"call it big Gamma,"},{"Start":"01:35.570 ","End":"01:37.790","Text":"which wraps all the way round."},{"Start":"01:37.790 ","End":"01:39.560","Text":"Here\u0027s the picture for it."},{"Start":"01:39.560 ","End":"01:43.070","Text":"You\u0027ll take it of radius 5 let say."},{"Start":"01:43.070 ","End":"01:47.330","Text":"This Gamma will use it to contradict the conditions"},{"Start":"01:47.330 ","End":"01:51.695","Text":"here and thereby show that f doesn\u0027t have an analytic log."},{"Start":"01:51.695 ","End":"01:56.485","Text":"You\u0027ll have to find one example of a Gamma where this integral is not 0."},{"Start":"01:56.485 ","End":"01:58.695","Text":"This will be our example,"},{"Start":"01:58.695 ","End":"02:02.595","Text":"0 and 4 are both inside Gamma."},{"Start":"02:02.595 ","End":"02:07.745","Text":"That means that if we let N and P be the number of zeros respectively poles,"},{"Start":"02:07.745 ","End":"02:09.950","Text":"then we have that N=2,"},{"Start":"02:09.950 ","End":"02:13.680","Text":"there are two zeros and no poles."},{"Start":"02:13.870 ","End":"02:16.730","Text":"N-P=2. Now by the argument principle,"},{"Start":"02:16.730 ","End":"02:22.250","Text":"this integral on our big Gamma =2 Pi i(N-P) comes out to be 4 Pi i."},{"Start":"02:22.250 ","End":"02:24.470","Text":"Anyway, the point is, it\u0027s not 0."},{"Start":"02:24.470 ","End":"02:27.335","Text":"Once we find one example which is not 0,"},{"Start":"02:27.335 ","End":"02:30.620","Text":"shows that f does not have an analytic logarithm."},{"Start":"02:30.620 ","End":"02:37.690","Text":"For part 2, we\u0027ll use the other theorem that if f has an analytic log on D,"},{"Start":"02:37.690 ","End":"02:43.160","Text":"then it also has an analytic nth root for all n positive integers,"},{"Start":"02:43.160 ","End":"02:45.425","Text":"and in particular for N=2."},{"Start":"02:45.425 ","End":"02:49.250","Text":"Now, I\u0027ll describe the general plan then fill in some of the details."},{"Start":"02:49.250 ","End":"02:54.645","Text":"For little f, we got that N-p was 2 not 0."},{"Start":"02:54.645 ","End":"02:58.590","Text":"In other words, N-P for the function f on the contour,"},{"Start":"02:58.590 ","End":"03:00.360","Text":"Gamma was not 0,"},{"Start":"03:00.360 ","End":"03:02.625","Text":"so that was no good."},{"Start":"03:02.625 ","End":"03:08.010","Text":"So what we\u0027re going to do, the trick is to write f as z^2g(z),"},{"Start":"03:08.010 ","End":"03:09.710","Text":"where g, well,"},{"Start":"03:09.710 ","End":"03:13.875","Text":"is just what you get if you divide f(z) by z^2."},{"Start":"03:13.875 ","End":"03:16.850","Text":"Here\u0027s a reminder of what f(z) was in case you forgot."},{"Start":"03:16.850 ","End":"03:19.665","Text":"So if we take z^2 out of it,"},{"Start":"03:19.665 ","End":"03:22.860","Text":"what we\u0027re left with is z minus 4 over z."},{"Start":"03:22.860 ","End":"03:25.830","Text":"Now for g, we show that N"},{"Start":"03:25.830 ","End":"03:34.200","Text":"minus P=0 along any contour Gamma in D for the function g. For the function f,"},{"Start":"03:34.200 ","End":"03:36.770","Text":"we\u0027ve got an example where it was 2."},{"Start":"03:36.770 ","End":"03:37.933","Text":"Once we\u0027ve shown this,"},{"Start":"03:37.933 ","End":"03:40.862","Text":"and I\u0027ll leave that for the details in a moment,"},{"Start":"03:40.862 ","End":"03:44.539","Text":"then we\u0027ll be able to say that this integral is"},{"Start":"03:44.539 ","End":"03:49.310","Text":"0 because it\u0027s 2 Pi i times N minus P and if N minus P is 0,"},{"Start":"03:49.310 ","End":"03:50.920","Text":"then this is 0."},{"Start":"03:50.920 ","End":"03:52.970","Text":"Then by the first theorem,"},{"Start":"03:52.970 ","End":"03:55.700","Text":"g will have an analytic logarithm in D,"},{"Start":"03:55.700 ","End":"03:58.040","Text":"and from the second theorem, therefore,"},{"Start":"03:58.040 ","End":"03:59.840","Text":"g has an analytic square root,"},{"Start":"03:59.840 ","End":"04:02.615","Text":"call it square root of g(z)."},{"Start":"04:02.615 ","End":"04:08.810","Text":"Then we can define h(z) to be z times the square root of g(z)."},{"Start":"04:08.810 ","End":"04:12.570","Text":"If we square h, we get z^2g(z),"},{"Start":"04:12.700 ","End":"04:17.260","Text":"and z^2g(z) is back to f(z)."},{"Start":"04:17.260 ","End":"04:19.415","Text":"That will be the square root."},{"Start":"04:19.415 ","End":"04:24.810","Text":"That will show that h(z) is square root of f(z) on D,"},{"Start":"04:24.810 ","End":"04:26.615","Text":"and so I have found an analytic square root."},{"Start":"04:26.615 ","End":"04:34.910","Text":"So we\u0027re left with showing that N minus P=0 on Gamma for g,"},{"Start":"04:34.910 ","End":"04:39.100","Text":"for any Gamma in D. This one I\u0027m saying,"},{"Start":"04:39.100 ","End":"04:40.790","Text":"so let\u0027s do that."},{"Start":"04:40.790 ","End":"04:46.315","Text":"Now, recall that g is equal to this maybe there it is,"},{"Start":"04:46.315 ","End":"04:49.290","Text":"you can still see it, so that\u0027s that."},{"Start":"04:49.290 ","End":"04:52.175","Text":"Now if we look in all of the complex plane,"},{"Start":"04:52.175 ","End":"04:58.865","Text":"then g has a 0 at 4 and a pole at 0."},{"Start":"04:58.865 ","End":"05:01.940","Text":"But neither of these is in D because remember D is"},{"Start":"05:01.940 ","End":"05:06.110","Text":"the exterior of the circle of radius 4."},{"Start":"05:06.110 ","End":"05:10.175","Text":"So g is non-zero and analytic."},{"Start":"05:10.175 ","End":"05:15.590","Text":"It has no poles in D. All we have to do to fill the conditions of"},{"Start":"05:15.590 ","End":"05:21.920","Text":"the theorem is to show that the integral is 0 along Gamma, this one here."},{"Start":"05:21.920 ","End":"05:26.060","Text":"Let Gamma be a simple closed contour in D. 0 and 4, I claim,"},{"Start":"05:26.060 ","End":"05:27.290","Text":"are either both inside,"},{"Start":"05:27.290 ","End":"05:29.600","Text":"Gamma or both outside Gamma,"},{"Start":"05:29.600 ","End":"05:32.455","Text":"we can\u0027t have one inside and one outside."},{"Start":"05:32.455 ","End":"05:34.910","Text":"Otherwise, lets see if I got a picture here."},{"Start":"05:34.910 ","End":"05:39.860","Text":"Yeah. If 0 was inside and 4 was outside,"},{"Start":"05:39.860 ","End":"05:42.230","Text":"then the curve Gamma,"},{"Start":"05:42.230 ","End":"05:46.370","Text":"the contour would cut some point along this segment."},{"Start":"05:46.370 ","End":"05:51.290","Text":"It can\u0027t go from inside to outside without cutting the curve and this is not in D,"},{"Start":"05:51.290 ","End":"05:53.300","Text":"so that would be a contradiction."},{"Start":"05:53.300 ","End":"05:55.610","Text":"This is one possibility."},{"Start":"05:55.610 ","End":"05:59.645","Text":"That\u0027s where both points are outside of Gamma."},{"Start":"05:59.645 ","End":"06:03.080","Text":"The other possibility is that they\u0027re both inside,"},{"Start":"06:03.080 ","End":"06:05.650","Text":"that we\u0027d get something like this."},{"Start":"06:05.650 ","End":"06:07.665","Text":"Now in either case,"},{"Start":"06:07.665 ","End":"06:09.765","Text":"we get N minus P=0,"},{"Start":"06:09.765 ","End":"06:13.245","Text":"because in this case, N=P=1."},{"Start":"06:13.245 ","End":"06:15.525","Text":"The pole and the 0 are both inside."},{"Start":"06:15.525 ","End":"06:18.340","Text":"In the other case, N=P=0."},{"Start":"06:18.340 ","End":"06:20.360","Text":"The pole and the 0 are both outside,"},{"Start":"06:20.360 ","End":"06:22.910","Text":"but in either case, N minus P=0."},{"Start":"06:22.910 ","End":"06:26.000","Text":"That\u0027s all we had to show to complete the proof,"},{"Start":"06:26.000 ","End":"06:28.770","Text":"and so we are done."}],"ID":23786}],"Thumbnail":null,"ID":102294}]

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