
Title: CauchyRiemann Equations Examples

Series: Complex Analysis

YouTubeTitle: Complex Analysis  Part 7  CauchyRiemann Equations Examples

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Timestamps
00:00 Intro 00:18 Quick recap 02:42 Example 1: Identity function 04:28 Example 2: Complex conjugate function 05:38 Example 3: Complex polynomial 
Subtitle in English
1 00:00:00,457 –> 00:00:03,626 Hello and welcome back to complex analysis.
2 00:00:04,586 –> 00:00:10,827 and first I want to thank all the very nice people that support this channel on Steady, via Paypal or by other means.
3 00:00:11,743 –> 00:00:17,131 Now, in today’s part 7 I show you examples for the famous CauchyRiemann equations.
4 00:00:18,100 –> 00:00:22,444 For this let’s first recall the important theorem from the last video.
5 00:00:22,800 –> 00:00:28,554 There we introduced the CauchyRiemann equations with respect to the complex differentiability.
6 00:00:29,343 –> 00:00:33,884 Hence we can immediately reformulate the theorem for holomorphic functions.
7 00:00:34,700 –> 00:00:39,798 and as always for this we have to choose an open subset capital U in C.
8 00:00:40,400 –> 00:00:47,649 and now you should already know the complex function can be equivalently described as a function from R^2 into R^2.
9 00:00:48,200 –> 00:00:54,424 However in this case for the new function the domain should also be an open subset of R^2.
10 00:00:55,014 –> 00:00:57,625 So maybe let’s simply call it U_R.
11 00:00:58,686 –> 00:01:04,217 Obviously when you visualize U and U_R in the plane, they look exactly the same.
12 00:01:04,971 –> 00:01:09,476 In other words the open domain here does not make any problems at all.
13 00:01:10,300 –> 00:01:17,689 Ok, now in the last video we have learned that we can split up the function f into a real and an imaginary part.
14 00:01:18,343 –> 00:01:24,493 and both we can see as a function from U_R as a subset of R^2 into R.
15 00:01:24,886 –> 00:01:29,903 In particular the real part function here, we denote by a lower case u.
16 00:01:30,414 –> 00:01:34,057 So please recall this u is a realvalued function.
17 00:01:34,971 –> 00:01:41,053 and then exactly in the same way, the imaginary part of f gives us a function we call v.
18 00:01:41,414 –> 00:01:46,513 So also here we have a function with domain U_R and codomain R.
19 00:01:47,443 –> 00:01:53,299 Hence you see, the whole information of the function f is now translated into these 2 functions.
20 00:01:54,071 –> 00:02:02,086 Indeed the property that f is holomorphic translates into the fact that u and v fulfill the CauchyRiemann equations.
21 00:02:03,014 –> 00:02:07,826 These are not complicated at all. There are just 2 partial differential equations.
22 00:02:08,557 –> 00:02:17,597 The first one simply says that the partial derivative of u with respect to x is the same as the partial derivative of v with respect to y.
23 00:02:18,100 –> 00:02:23,286 and then in the second equation you see, when you switch the roles there is a minus sign included.
24 00:02:24,386 –> 00:02:32,388 In summary what we need for a holomorphic function is that these 2 equations are fulfilled at each point in the set U_R.
25 00:02:33,457 –> 00:02:37,523 In fact this equivalence here, we have discussed in the last video.
26 00:02:38,043 –> 00:02:41,611 Now in this video I want to show you concrete examples.
27 00:02:42,400 –> 00:02:46,033 and I would say, let’s immediately start with a very simple one.
28 00:02:46,986 –> 00:02:53,767 So let’s take the complex function f from C to C. Given as the identity.
29 00:02:54,671 –> 00:02:57,951 Which means the number z is simply mapped to z again.
30 00:02:59,014 –> 00:03:06,656 Now, because we want to deal with the partial derivatives with respect to x and y, we have to write z as (x + iy).
31 00:03:07,786 –> 00:03:11,431 Hence we do this change on the lefthand side and on the righthand side.
32 00:03:12,314 –> 00:03:19,045 and then you see, here the real part is our function u and the imaginary part is our function v.
33 00:03:19,929 –> 00:03:23,830 So you see in this case it’s not complicated at all.
34 00:03:24,500 –> 00:03:28,929 and now we just want to check if the CauchyRiemann equations are satisfied.
35 00:03:29,957 –> 00:03:32,489 So let’s write down the partial derivatives.
36 00:03:33,300 –> 00:03:37,928 First we have du/dx. Which is in this case simply 1.
37 00:03:38,729 –> 00:03:47,597 In the same way we also see that dv/dy is very easy to calculate, because this is a simple function and the derivative is just 1 again.
38 00:03:48,771 –> 00:03:53,901 Hence we immediately see that the first equation is satisfied at all points.
39 00:03:54,271 –> 00:03:58,434 So let’s go to the second equation, where we have du/dy.
40 00:03:59,514 –> 00:04:03,578 There we see, there is no y in u. Therefore we get 0.
41 00:04:04,271 –> 00:04:08,837 In fact the same holds for dv/dx, because there is no x in v.
42 00:04:09,371 –> 00:04:12,459 and also of course, the minus sign does not change anything.
43 00:04:13,600 –> 00:04:17,681 In summary, both CauchyRiemann equations are fulfilled.
44 00:04:18,486 –> 00:04:22,290 From that we then can conclude that the function f is holomorphic.
45 00:04:23,429 –> 00:04:26,046 Of course this is a fact we have already known.
46 00:04:26,857 –> 00:04:33,745 Therefore I would say in the next example, let’s look at a function where we already know that it is not holomorphic.
47 00:04:34,571 –> 00:04:37,996 Indeed this was simply the complex conjugation.
48 00:04:38,814 –> 00:04:40,470 So f(z) is z bar.
49 00:04:41,500 –> 00:04:47,868 Now as before, in order to apply the CauchyRiemann equations we have to rewrite that with x and y.
50 00:04:48,586 –> 00:04:53,698 Of course this is not so complicated. The imaginary part just gets a minus sign.
51 00:04:54,743 –> 00:04:59,862 and then as before, the real part is u and the imaginary part here is v.
52 00:05:00,471 –> 00:05:05,664 So you see it’s different from before and it will change the CauchyRiemann equation.
53 00:05:06,771 –> 00:05:13,054 Of course here du/dx is still 1, but dv/dy is now 1.
54 00:05:13,914 –> 00:05:17,401 Therefore we can conclude: this is not the same.
55 00:05:18,057 –> 00:05:20,301 Indeed no matter which point we put in.
56 00:05:21,043 –> 00:05:26,498 In conclusion we get the result, we’ve expected. Namely f is not holomorphic.
57 00:05:27,500 –> 00:05:33,013 Here you see, showing this was very simple by just using the CauchyRiemann equations.
58 00:05:33,943 –> 00:05:38,022 However maybe we should also look at a more complicated example.
59 00:05:39,014 –> 00:05:42,099 So let’s take another complex function f.
60 00:05:42,814 –> 00:05:47,326 and now I want to look at it as z^2 + iz.
61 00:05:47,871 –> 00:05:54,400 To be honest. You already know that this is a holomorphic function, because it is a complex polynomial.
62 00:05:55,200 –> 00:06:01,485 Indeed we already know how to calculate the complex derivative for such a complex polynomial.
63 00:06:02,257 –> 00:06:05,101 Here it should be 2 times z + i.
64 00:06:05,543 –> 00:06:09,531 Still it’s helpful to check it with the CauchyRiemann equations.
65 00:06:10,586 –> 00:06:16,479 Now, by writing x + iy instead of z, we actually have something to do now,
66 00:06:17,057 –> 00:06:23,119 because in this expression here, we don’t see immediately the real and imaginary part.
67 00:06:24,100 –> 00:06:26,876 We first have to calculate a little bit.
68 00:06:27,500 –> 00:06:32,139 However that’s not so complicated, we simply can do the whole calculation here.
69 00:06:32,614 –> 00:06:42,871 So we have x^2 + i2xy  y^2 from the binomial + ix and y.
70 00:06:43,729 –> 00:06:48,299 Now let’s put the real part to the left and the imaginary part to the right.
71 00:06:49,014 –> 00:06:55,322 So the real part is everything without an “i” and for the imaginary part I factor out one “i”.
72 00:06:56,457 –> 00:06:59,190 and then we have the result we wanted,
73 00:07:00,057 –> 00:07:05,038 because again, here we have our function u and there we have the function v.
74 00:07:05,986 –> 00:07:10,214 Now with these 2 functions we can check the CauchyRiemann equations again.
75 00:07:11,386 –> 00:07:15,814 First du/dx, not hard to see, is simply 2 times x
76 00:07:16,800 –> 00:07:23,738 Then the second partial derivative we want to calculate is dv/dy, which is also 2x.
77 00:07:24,857 –> 00:07:28,398 Hence our first equation is actually fulfilled.
78 00:07:29,000 –> 00:07:37,122 Going to the second one, you see we have to do a little bit more, because this derivative is 2y  1
79 00:07:37,714 –> 00:07:44,592 Ok, then the last derivative we want to calculate would be the partial derivative of v with respect to x.
80 00:07:45,457 –> 00:07:48,741 and there you see, it’s 2y + 1
81 00:07:49,543 –> 00:07:53,746 and then for our equation we just have to put a minus in front.
82 00:07:54,229 –> 00:07:58,317 and then we recognize this is the same as du/dy.
83 00:07:59,071 –> 00:08:04,858 Therefore also here both equations are fulfilled. So the function f is holomorphic.
84 00:08:05,514 –> 00:08:08,421 I already told you, this is not a surprise for us.
85 00:08:09,429 –> 00:08:14,871 However with this example you have learned how you can apply the CauchyRiemann equations,
86 00:08:14,914 –> 00:08:18,857 when the function f is given as a real and an imaginary part.
87 00:08:19,820 –> 00:08:25,952 and in fact you also get the complex derivative of f from these partial derivatives here,
88 00:08:26,757 –> 00:08:30,069 but I think this is a good topic for the next video.
89 00:08:30,757 –> 00:08:36,457 Therefore I really hope that I see you there. Have a nice day and bye!