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Title: Cauchy-Riemann Equations
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Series: Complex Analysis
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YouTube-Title: Complex Analysis 6 | Cauchy-Riemann Equations
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Bright video: https://youtu.be/BvvFdmOrob4
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Dark video: https://youtu.be/RG-letVCC9Q
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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: ca06_sub_eng.srt
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Timestamps
00:00 Intro
00:16 Two notions of differentiability
06:39 When is the vector-matrix multiplication a complex multiplication?
09:04 Deriving the Cauchy-Riemann equations
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Subtitle in English
1 00:00:00,714 –> 00:00:04,186 Hello and welcome back to complex analysis.
2 00:00:05,214 –> 00:00:13,547 and you might already know, first as always I want to thank all the nice people that support this channel on Steady, via Paypal or by other means.
3 00:00:14,500 –> 00:00:19,505 Now, finally in today’s part 6, we will talk about the Cauchy-Riemann equations.
4 00:00:20,400 –> 00:00:28,511 In order to understand them we first have to recall the 2 notions of differentiability we have for complex function.
5 00:00:29,500 –> 00:00:34,875 The first is the normal one. The one we introduced at the start of this series.
6 00:00:35,729 –> 00:00:40,500 This one we simply called complex differentiable at a given point z_0.
7 00:00:41,400 –> 00:00:49,303 Now, as a reminder in other videos I already explained, the domain here doesn’t have to be the whole complex plane C.
8 00:00:50,043 –> 00:00:52,163 It’s sufficient to have an open subset.
9 00:00:53,271 –> 00:00:59,414 However this might be distracting for this video here. Therefore I always write C for the domain.
10 00:01:00,186 –> 00:01:05,593 So if needed you can just substitute this domain with any open subset U.
11 00:01:06,471 –> 00:01:13,330 Ok, back to the differentiability here. This one can be described as a linear approximation in C.
12 00:01:13,614 –> 00:01:20,628 More precisely this means that we can find the complex number, we can call f’(z_0).
13 00:01:21,371 –> 00:01:24,386 Indeed, this could be any complex number.
14 00:01:25,457 –> 00:01:31,913 Then with this it’s possible for us to write f(z) as a linear term + an error term.
15 00:01:32,629 –> 00:01:39,361 and you already know, everything happens locally around z_0. Therefore the constant term is f(z_0).
16 00:01:40,429 –> 00:01:46,243 Then next comes the linear map. Which has a slope given by f’(z_0).
17 00:01:46,957 –> 00:01:51,685 More precisely we have this number times (z - z_0).
18 00:01:52,757 –> 00:01:58,569 So you see again, this is the linear approximation of the function f around the point z_0.
19 00:01:59,314 –> 00:02:03,637 and please recall, in real analysis this represents the tangent.
20 00:02:04,300 –> 00:02:09,557 Therefore the only thing missing here is a corresponding error term, we could call Phi.
21 00:02:10,114 –> 00:02:14,490 Of course this Phi is also a function from C to C.
22 00:02:15,129 –> 00:02:21,782 Now as an error function this map here should go to 0, when we send z to z_0.
23 00:02:22,686 –> 00:02:28,086 and this convergence to 0 should go faster to 0 than the linear function would do.
24 00:02:28,986 –> 00:02:39,600 Or more concretely Phi(z) divided by (z - z_0) should still go to 0, when we send z to z_0.
25 00:02:40,414 –> 00:02:46,112 Now, with this property we really get a linear approximation around the point z_0.
26 00:02:46,771 –> 00:02:53,583 Indeed, this is what we could read as the definition for complex differentiability at the point z_0.
27 00:02:54,243 –> 00:02:58,955 However to be honest it looks a little bit different than the definition we gave before.
28 00:02:59,786 –> 00:03:05,438 Yet I would say it’s not hard at all to get this reformulation from our definition.
29 00:03:06,143 –> 00:03:09,651 and I guess it’s a good exercise for you to think about it.
30 00:03:10,114 –> 00:03:16,099 After this you see, you simply can take this for the definition of complex differentiability.
31 00:03:16,757 –> 00:03:25,328 and that’s a good thing, because we can nicely compare this to the definition of total differentiability we explained in the last video.
32 00:03:26,000 –> 00:03:31,869 As a reminder, this definition we introduced for functions from R^2 into R^2.
33 00:03:32,571 –> 00:03:40,739 and there please recall, we have a 1 to 1 correspondence between functions from C to C and functions from R^2 into R^2.
34 00:03:41,471 –> 00:03:45,929 Therefore often both functions are just called by the same name, f.
35 00:03:46,426 –> 00:03:51,028 However to distinguish them I give the last one here an index R.
36 00:03:51,657 –> 00:03:57,547 Ok, now we can recall the definition of total differentiability for such functions.
37 00:03:58,443 –> 00:04:03,571 Also there we have to fix a point in the plane. So let’s call it (x_0, y_0).
38 00:04:04,171 –> 00:04:12,446 Then we know the function is called totally differentiable at this point, if there exists a matrix J and an error map Phi.
39 00:04:13,014 –> 00:04:16,817 Please note I use another variant to denote the letter Phi here.
40 00:04:17,829 –> 00:04:23,406 Then what we need is also a linear approximation. Now given by this matrix J.
41 00:04:24,186 –> 00:04:30,114 and lastly of course this error term here should also go to 0 in the same way as before.
42 00:04:30,786 –> 00:04:37,255 Which means if we divide it by the length of the difference vector, we still get out 0 in this limit.
43 00:04:37,914 –> 00:04:45,588 Here please note, before this 0 was the 0 in the complex numbers and now this 0, is the 0 vector in R^2.
44 00:04:46,286 –> 00:04:52,723 However using the identification C with R^2, you could say they are exactly the same.
45 00:04:53,500 –> 00:04:59,800 Therefore a natural question here would be: where exactly is the difference between both notions here?
46 00:05:00,500 –> 00:05:04,329 So I would say let’s try to spot all the differences.
47 00:05:05,343 –> 00:05:13,694 First this might be obvious, in the second case we see when we use the division here, we need the length of the vector.
48 00:05:14,543 –> 00:05:17,585 This is what you know, we can’t divide vectors.
49 00:05:18,514 –> 00:05:26,071 However in the complex numbers we don’t have this problem, we simply can divide one complex number by another one.
50 00:05:26,871 –> 00:05:34,610 Now, maybe this is not a crucial difference, because both things just explain that the error term goes to 0.
51 00:05:35,571 –> 00:05:39,895 and of course if we wanted, we also could use the absolute value here.
52 00:05:40,743 –> 00:05:44,486 Indeed this wouldn’t change anything for the linear approximation here.
53 00:05:45,300 –> 00:05:50,424 Therefore we have to look at the linear approximations to see the differences.
54 00:05:51,443 –> 00:05:56,977 Since we immediately see that the constant terms fit, we look at the linear terms here.
55 00:05:57,800 –> 00:06:04,043 and there you see, this dot here stands for the multiplication in the complex numbers C.
56 00:06:04,943 –> 00:06:10,080 and that’s the difference because this multiplication does not exist here.
57 00:06:10,743 –> 00:06:15,231 Indeed what we find here is a matrix-vector multiplication.
58 00:06:16,229 –> 00:06:22,157 Now exactly this does not translate in general to the multiplication in C.
59 00:06:22,771 –> 00:06:31,714 Therefore the conclusion is immediately this definition here, complex differentiability is much stricter than this definition there.
60 00:06:32,586 –> 00:06:40,921 and now in order to get a translation from the complex definition to the one in R^2, we have to answer one question.
61 00:06:41,786 –> 00:06:49,812 Namely in which cases does a matrix-vector multiplication represent a multiplication in the complex numbers.
62 00:06:50,829 –> 00:06:57,609 Now, because we know how to calculate with complex numbers, we can answer this question without any problem.
63 00:06:58,586 –> 00:07:02,932 Just let’s check how the multiplication in the complex numbers looks like.
64 00:07:03,814 –> 00:07:08,497 So we take 2 complex numbers w and z, and multiply them.
65 00:07:09,271 –> 00:07:15,062 Of course here w and z have a real part and an imaginary part respectively.
66 00:07:15,786 –> 00:07:22,438 Therefore we say w = (a + ib) and z = (x + iy).
67 00:07:23,443 –> 00:07:28,495 and now we just do the multiplication, where we use that i^2 is -1.
68 00:07:29,071 –> 00:07:36,209 Then we get: a times x - b times y + the imaginary part.
69 00:07:37,043 –> 00:07:40,568 Which is b times x + a times y.
70 00:07:41,386 –> 00:07:46,793 Of course this was not hard at all, because you know how to calculate with complex numbers.
71 00:07:47,686 –> 00:07:52,514 However now we want to rewrite this as a matrix-vector multiplication.
72 00:07:53,014 –> 00:07:59,516 and the vector should represent the complex number z. Which means we have x and y as the coordinates.
73 00:08:00,314 –> 00:08:05,601 and on the left of it, we have a 2x2 matrix where the numbers a and b should occur.
74 00:08:06,486 –> 00:08:10,566 and indeed how the entries exactly look like we have to figure out now.
75 00:08:11,571 –> 00:08:15,889 and of course this is no problem for us, because we already know the result.
76 00:08:16,771 –> 00:08:20,155 The first component should just be the real part here.
77 00:08:20,814 –> 00:08:24,285 and then the second component is just the imaginary part.
78 00:08:25,286 –> 00:08:30,302 and now the question is: Which 2x2 matrix brings us to this result?
79 00:08:31,071 –> 00:08:35,835 Now the first row we see immediately, because it should be “a” and -b.
80 00:08:36,657 –> 00:08:41,982 Of course in the same way we also see the second row. Which be b and a.
81 00:08:42,571 –> 00:08:48,656 and indeed that’s it. This is the matrix that represents the multiplication of complex numbers.
82 00:08:49,514 –> 00:08:58,084 Therefore our conclusion here is, if we want complex differentiability we need that the Jacobian matrix here has this form.
83 00:08:59,014 –> 00:09:03,505 So I would say let’s fix this important result with a theorem.
84 00:09:04,400 –> 00:09:09,014 In fact in this theorem we will now find the Cauchy-Riemann equations.
85 00:09:09,814 –> 00:09:14,381 Ok in order to get everything together let’s start with a complex function.
86 00:09:14,414 –> 00:09:18,457 That should be complex differentiable at a given point z_0.
87 00:09:19,357 –> 00:09:23,571 and now we can just split this point into a real and an imaginary part.
88 00:09:24,414 –> 00:09:28,901 and as before we call them x_0 and y_0 respectively.
89 00:09:29,729 –> 00:09:34,722 So with the reasoning from above we know that this statement here is equivalent to
90 00:09:35,743 –> 00:09:44,798 The statement that the corresponding real function f_R from R^2 to R^2 is totally differentiable at the point (x_0, y_0)
91 00:09:45,157 –> 00:09:52,043 and that the Jacobian matrix at the point (x_0, y_0) has the symmetric form from above.
92 00:09:53,143 –> 00:09:58,571 So there you see this is our final connection between both notions of differentiability.
93 00:09:59,600 –> 00:10:04,437 and exactly this connection now leads us to the Cauchy-Riemann equations.
94 00:10:05,343 –> 00:10:12,501 We simply get another equivalence when we use the fact that the Jacobian matrix has partial derivatives as entries.
95 00:10:13,329 –> 00:10:19,543 However in order to do this we have to introduce new names for the components of the function f_R.
96 00:10:20,357 –> 00:10:24,214 That’s no problem at all, because we know we just have 2 components.
97 00:10:25,229 –> 00:10:29,036 The first one we simply call u and the second one v.
98 00:10:30,071 –> 00:10:33,963 Now both components depend on the 2 variables x and y.
99 00:10:34,557 –> 00:10:38,491 Therefore we simply write u(x, y) and v(x, y).
100 00:10:39,529 –> 00:10:45,015 In other words here we have 2 new maps that send R^2 to R.
101 00:10:45,971 –> 00:10:49,118 So please don’t forget, here we just have real functions.
102 00:10:49,914 –> 00:10:58,291 Ok and now we know from the last video that in the top left corner for example we find the partial derivative of u with respect to x.
103 00:10:59,257 –> 00:11:04,807 and now you see, with the partial derivatives in the Jacobian matrix we find a connection.
104 00:11:05,729 –> 00:11:10,069 And exactly this connection is what we call the Cauchy-Riemann equations.
105 00:11:11,057 –> 00:11:15,141 There are just 2 partial equations for u and v.
106 00:11:15,871 –> 00:11:24,154 In particular the first one is simply that du/dx = dv/dy.
107 00:11:25,300 –> 00:11:32,454 So here you see this equation just represents that we have the same number in the top left corner and in the bottom right corner.
108 00:11:33,200 –> 00:11:38,438 Ok and then you might already see it. The number b here gives us the second equation.
109 00:11:39,186 –> 00:11:44,211 Which is du/dy = dv/dx.
110 00:11:45,014 –> 00:11:50,604 and of course both equations have to be fulfilled at the given point (x_0, y_0).
111 00:11:51,414 –> 00:11:59,050 Ok now you know that we have these 2 equivalences to describe complex differentiability at a given point.
112 00:11:59,743 –> 00:12:06,205 and of course then we can extend that to a whole open subset. We talk about holomorphic functions.
113 00:12:06,986 –> 00:12:13,468 In other words a holomorphic function has to fulfill the Cauchy-Riemann equations at all the points.
114 00:12:14,114 –> 00:12:16,514 Ok then i think it’s good enough for today.
115 00:12:16,613 –> 00:12:19,745 Let’s talk about examples in the next video.
116 00:12:20,343 –> 00:12:24,211 Therefore I hope I see you there and have a nice day. Bye!
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Quiz Content
Q1: Consider the complex function $f: \mathbb{C} \rightarrow \mathbb{C}$ given by $f(z) = z^2 + i z$. The function holomorphic because it is…
A1: a complex polynomial.
A2: a function with domain $\mathbb{C}$.
A3: a real-valued function.
A4: a linear function.
Q2: Consider again the complex function $f: \mathbb{C} \rightarrow \mathbb{C}$ given by $f(z) = z^2 + i z$. What is the corresponding map $f_R: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with $$f\left( \binom{x}{y} \right) = \binom{ u(x,y) }{ v(x,y) }$$
A1: $u(x,y) = x^2 - y^2$ and $v(x,y) = 2 x y$
A2: $u(x,y) = x^2 $ and $v(x,y) = 2 x $
A3: $u(x,y) = x^2 - y^2 - y$ and $v(x,y) = 2 x y$
A4: $u(x,y) = x^2 - y^2 - y$ and $v(x,y) = 2 x y + x$
A5: $u(x,y) = x^2 - y^2 - y^2$ and $v(x,y) = 2 x y + x^2$
A7: $u(x,y) = x^2 - y^2 - 2 y$ and $v(x,y) = 2 x y+ x^2$
Q3: What are the correct Cauchy-Riemann equations?
A1: $ \frac{\partial u}{ \partial x} = \frac{\partial v}{ \partial x}$ and $ \frac{\partial u}{ \partial y} = \frac{\partial v}{ \partial y}$
A2: $ \frac{\partial u}{ \partial x} = - \frac{\partial v}{ \partial y}$ and $ \frac{\partial u}{ \partial y} = \frac{\partial v}{ \partial x}$
A3: $ \frac{\partial u}{ \partial x} = \frac{\partial v}{ \partial x}$ and $ \frac{\partial u}{ \partial y} = -\frac{\partial v}{ \partial y}$
A4: $ \frac{\partial u}{ \partial x} = \frac{\partial v}{ \partial y}$ and $ \frac{\partial u}{ \partial y} = - \frac{\partial v}{ \partial x}$
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Last update: 2024-10
