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Title: Wirtinger Derivatives
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Series: Complex Analysis
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YouTube-Title: Complex Analysis 8 | Wirtinger Derivatives
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Bright video: https://youtu.be/5tJ-EDx7MlM
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Dark video: https://youtu.be/l1yJmTCoylM
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Ad-free video: Watch Vimeo video
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: ca08_sub_eng.srt
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Timestamps
00:00 Intro
00:19 Wirtinger derivatives — basic definition
01:35 Complex derivative through Cauchy-Riemann equations
05:12 Wirtinger derivatives — detailed definition
06:17 Example for z²
08:39 Summary: a criteria for holomorphic functions
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Subtitle in English
1 00:00:00,543 –> 00:00:03,958 Hello and welcome back to complex analysis.
2 00:00:04,671 –> 00:00:11,221 and as always, first I want to thank all the nice people that support this channel on Steady, via Paypal or by other means.
3 00:00:12,014 –> 00:00:20,171 Today in part 8 we will continue talking about the Cauchy-Riemann equations and I will also explain what Wirtinger derivatives are.
4 00:00:20,786 –> 00:00:27,890 For this as often, we consider a complex function f, that is holomorphic on an open domain U.
5 00:00:28,771 –> 00:00:36,862 and for such a function we will define a new differential operator as a partial derivative with respect to the variable z.
6 00:00:37,614 –> 00:00:45,449 Then when we apply this to the function f and look at a given point z_0, this is often called the Wirtinger derivative.
7 00:00:46,414 –> 00:00:51,694 However, this is not quite correct, because there is also another Wirtinger derivative.
8 00:00:52,371 –> 00:00:59,075 This one is a little bit strange, because it’s written as the partial derivative with respect to z bar.
9 00:00:59,700 –> 00:01:06,654 However it turns out that both things here are well defined and indeed useful for our theory.
10 00:01:07,214 –> 00:01:15,197 For example: for a holomorphic function f this one here should give us the complex derivative of f at the point z_0.
11 00:01:15,771 –> 00:01:20,477 With this in mind, we will be able to define this new differential operator.
12 00:01:21,543 –> 00:01:28,172 Also, then we will see that for a holomorphic function, the other Wirtinger derivative should be 0.
13 00:01:28,986 –> 00:01:35,027 Therefore let’s first concentrate at the Wirtinger derivative on the left hand side. df/dz.
14 00:01:35,614 –> 00:01:42,105 Now, to make our life a little bit easier let’s write (x + iy) instead of the point z_0.
15 00:01:42,771 –> 00:01:47,621 So this means, this is the complex derivative of f at the given point.
16 00:01:47,821 –> 00:01:53,388 Hence it’s just another complex number, which we can write as (a + ib).
17 00:01:54,571 –> 00:02:02,194 and now this is what we have learned in the last videos. The real part and the imaginary part can be written as partial derivatives.
18 00:02:03,229 –> 00:02:10,960 This works when we take the real part of the function f as a function u and the imaginary part of f as a function v.
19 00:02:11,900 –> 00:02:18,438 and then this “a” here is du/dx and b is dv/dx.
20 00:02:19,743 –> 00:02:26,191 Of course both partial derivatives here are to be evaluated at the given point (x, y).
21 00:02:26,391 –> 00:02:32,450 However in the upcoming calculation we will omit this point to make everything clearer.
22 00:02:33,357 –> 00:02:38,070 Of course now in the next step here we want to use the Cauchy-Riemann equations.
23 00:02:39,271 –> 00:02:45,297 and in order to include them correctly the best thing we can do is to double this complex number here.
24 00:02:45,943 –> 00:02:51,519 So you see, we add the same number again, but then we have to divide everything by 2.
25 00:02:52,700 –> 00:02:57,994 and now the idea is, that here in the second part we apply the Cauchy-Riemann equations.
26 00:02:58,971 –> 00:03:03,171 So instead of du/dx we can write dv/dy.
27 00:03:03,671 –> 00:03:09,396 and on the other hand dv/dx can be written as -du/dy.
28 00:03:10,271 –> 00:03:18,933 Ok, what you now should see is that some symmetry is involved. We have 2 partial derivatives with respect to x and 2 with respect to y.
29 00:03:19,771 –> 00:03:27,665 and this is what we can apply to our knowledge that u is the real part of the function f and v the imaginary part.
30 00:03:28,429 –> 00:03:35,661 Or in other words we can also look at the map that has (x, y) as an input, but a complex number as an output.
31 00:03:36,314 –> 00:03:40,795 Namely f at the point (x + iy) should be the output.
32 00:03:41,243 –> 00:03:47,834 So you see as the map f_R this carries exactly the same information as the map f.
33 00:03:48,657 –> 00:03:51,139 It’s just again another point of u.
34 00:03:51,686 –> 00:03:57,282 and this helps us, because this function we can write as u + iv.
35 00:03:57,482 –> 00:04:02,848 and now you might already see, on the left-hand side we use exactly this.
36 00:04:03,343 –> 00:04:10,580 Here the first part we can write as the partial derivative with respect to x of u + iv.
37 00:04:10,971 –> 00:04:17,043 and now in order to get this in the second part as well you see we have to pull out a -i
38 00:04:17,671 –> 00:04:25,499 and when we do this, you see we also have the partial derivative now with respect to y of u + iv.
39 00:04:26,871 –> 00:04:31,038 Ok, so this is the result for f’ you really should remember.
40 00:04:31,614 –> 00:04:38,208 Indeed it’s easier to remember when we use that (u + iv) are essentially the function f.
41 00:04:38,814 –> 00:04:43,110 Hence we can just define 2 new differential operators.
42 00:04:44,143 –> 00:04:49,773 Namely we just call the first one df/dx and the second one df/dy.
43 00:04:50,343 –> 00:04:57,966 and indeed this makes sense. This one is the partial derivative with respect to x, when you see f as this map here.
44 00:04:58,729 –> 00:05:04,143 Ok, soon I will show you an example and then these 2 operations should be clear.
45 00:05:04,771 –> 00:05:12,456 However first I want to define the Wirtinger derivatives, because with this calculation here we already know how to do this.
46 00:05:13,071 –> 00:05:18,456 We can just use combinations of the partial derivatives with respect to x and y.
47 00:05:19,257 –> 00:05:29,814 Firstly, now when we write d/dz. This should stand for 1/2 times d/dx - i*d/dy.
48 00:05:30,786 –> 00:05:34,514 So you see, this is exactly what we have learned above.
49 00:05:35,171 –> 00:05:41,471 The definition makes sense and for a holomorphic function f it gives us the complex derivative of f.
50 00:05:42,357 –> 00:05:47,022 Ok, now in a similar way we will define d/dz(bar).
51 00:05:48,357 –> 00:05:56,244 and indeed this should look very similarly. So we also take 1/2 times, and now d/dx + i d/dy.
52 00:05:56,943 –> 00:06:06,232 We do this, because you see by the Cauchy-Riemann equations, we get 0 when we apply this operation to a holomorphic function f.
53 00:06:07,243 –> 00:06:11,469 and here please recall. This fits to our motivation from the beginning.
54 00:06:11,929 –> 00:06:15,627 Ok, then as promised let’s look at an example.
55 00:06:16,071 –> 00:06:19,976 It shouldn’t be a complicated one. So let’s look at a polynomial.
56 00:06:20,671 –> 00:06:24,550 Namely we just look at a quadratic function z^2.
57 00:06:24,957 –> 00:06:29,848 Which means using x and y, we have (x + iy)^2.
58 00:06:30,500 –> 00:06:41,763 Hence, let’s just expand the square. Which means we have x^2 - y^2 + i2xy.
59 00:06:42,986 –> 00:06:48,463 Therefore this first part here is our function u and this part is our function v.
60 00:06:49,029 –> 00:06:57,384 With this we can immediately calculate df/dx, which is here 2x + i2y.
61 00:06:58,014 –> 00:07:06,883 and similarly we can calculate df/dy, which is here -2y + i2x.
62 00:07:07,629 –> 00:07:15,551 Now, both things can be simplified when we just use the fact, that (x+iy) is the complex number z.
63 00:07:16,243 –> 00:07:20,939 Hence, df/dx is simply 2 times z.
64 00:07:21,857 –> 00:07:27,434 and in a similar way you see for df/dy we have 2 times i times z.
65 00:07:28,086 –> 00:07:31,638 So we can always use that i^2 is -1.
66 00:07:32,714 –> 00:07:39,054 Now, with these 2 nice short expressions we should be able to calculate with 2 Wirtinger derivatives.
67 00:07:40,043 –> 00:07:43,993 So let’s start with the second one, df/dz(bar).
68 00:07:44,314 –> 00:07:51,736 It’s 1/2 (2z + i times 2iz).
69 00:07:52,571 –> 00:07:57,384 Again, we know i^2 is -1. So we get 0 here.
70 00:07:58,229 –> 00:08:02,568 This is what we expected, because the polynomial is a holomorphic function.
71 00:08:03,300 –> 00:08:07,491 Ok, then in the next step it’s not hard to calculate df/dz.
72 00:08:07,800 –> 00:08:11,955 It’s the same calculation as before, but now with a minus sign here.
73 00:08:12,700 –> 00:08:21,545 Therefore in this case, in the parentheses we have (2z + 2z) divided by 1/2, gives us back 2z.
74 00:08:22,643 –> 00:08:29,352 Also here the result is not so surprising, because it should give us the complex derivative of f.
75 00:08:29,552 –> 00:08:36,384 Ok, then I would say let’s summarize the video by stating the important fact we have proven here.
76 00:08:36,886 –> 00:08:40,505 Indeed, this is something one can nicely remember.
77 00:08:41,486 –> 00:08:53,967 Namely, a complex function f defined on an open domain U is holomorphic if and only if df/dz(bar) is equal to 0 at all points.
78 00:08:54,471 –> 00:09:01,291 Indeed, this equivalence you already know, because here we simply have hidden the Cauchy-Riemann equations.
79 00:09:02,214 –> 00:09:10,338 and then in this case as we have proven above the other Wirtinger derivative gives us the complex derivative of f.
80 00:09:11,071 –> 00:09:15,370 So f’ can be calculated by using df/dz.
81 00:09:16,100 –> 00:09:22,701 Ok, with this you now know a lot of different differential operators we can use in complex analysis.
82 00:09:23,714 –> 00:09:28,420 Then I hope that I see you in the next video, where we will talk about power series.
83 00:09:29,557 –> 00:09:31,594 Have a nice day! Bye!
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Quiz Content
Q1: A complex function $f: \mathbb{C} \rightarrow \mathbb{C}$ is holomorphic if and only if
A1: $$\frac{ \partial f}{\partial z} = 0$$
A2: $$\frac{ \partial f}{\partial \overline{z}} = 0$$
A3: $$\frac{ \partial f}{\partial \overline{z}} = \frac{ \partial f}{\partial \overline{z}}$$
A4: $$\frac{ \partial f}{\partial \overline{z}} = 1$$
Q2: Consider the complex function $f: \mathbb{C} \rightarrow \mathbb{C}$ given by $f(z) = \overline{z} z $. Which claim is not correct?
A1: $$\frac{ \partial f}{\partial \overline{z}}(z) = z$$
A2: $$\frac{ \partial f}{\partial z}(z) = \overline{z}$$
A3: $$\frac{ \partial f}{\partial z}(z) = \overline{z} z$$
A4: $$\frac{ \partial f}{\partial z}(0) = 0$$
A5: $$\frac{ \partial f}{\partial \overline{z}}(0) = 0$$
Q3: Is the following correct? $$ \frac{ \partial f}{\partial z} = \overline{\frac{ \partial \overline{f}}{\partial \overline{z}}}$$
A1: Yes!
A2: No!
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Last update: 2024-10
