1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
|
---
title: "Holomorphic function"
date: 2021-02-25
categories:
- Mathematics
- Complex analysis
layout: "concept"
---
In complex analysis, a complex function $f(z)$ of a complex variable $z$
is called **holomorphic** or **analytic** if it is complex differentiable in the
neighbourhood of every point of its domain.
This is a very strong condition.
As a result, holomorphic functions are infinitely differentiable and
equal their Taylor expansion at every point. In physicists' terms,
they are extremely "well-behaved" throughout their domain.
More formally, a given function $f(z)$ is holomorphic in a certain region
if the following limit exists for all $z$ in that region,
and for all directions of $\Delta z$:
$$\begin{aligned}
\boxed{
f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
}
\end{aligned}$$
We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$:
$$\begin{aligned}
f(z) = f(x + i y) = u(x, y) + i v(x, y)
\end{aligned}$$
Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$:
$$\begin{aligned}
f'(z)
&= \lim_{\Delta x \to 0} \frac{f(z + \Delta x) - f(z)}{\Delta x}
= \pdv{u}{x} + i \pdv{v}{x}
\\
&= \lim_{\Delta y \to 0} \frac{f(z + i \Delta y) - f(z)}{i \Delta y}
= \pdv{v}{y} - i \pdv{u}{y}
\end{aligned}$$
For $f(z)$ to be holomorphic, these two results must be equivalent.
Because $u$ and $v$ are real by definition,
we thus arrive at the **Cauchy-Riemann equations**:
$$\begin{aligned}
\boxed{
\pdv{u}{x} = \pdv{v}{y}
\qquad
\pdv{v}{x} = - \pdv{u}{y}
}
\end{aligned}$$
Therefore, a given function $f(z)$ is holomorphic if and only if its real
and imaginary parts satisfy these equations. This gives an idea of how
strict the criteria are to qualify as holomorphic.
## Integration formulas
Holomorphic functions satisfy **Cauchy's integral theorem**, which states
that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero,
provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$:
$$\begin{aligned}
\boxed{
\oint_C f(z) \dd{z} = 0
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-int-theorem"/>
<label for="proof-int-theorem">Proof</label>
<div class="hidden">
<label for="proof-int-theorem">Proof.</label>
Just like before, we decompose $f(z)$ into its real and imaginary parts:
$$\begin{aligned}
\oint_C f(z) \dd{z}
&= \oint_C (u + i v) \dd{(x + i y)}
= \oint_C (u + i v) \:(\dd{x} + i \dd{y})
\\
&= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
\end{aligned}$$
Using Green's theorem, we integrate over the area $A$ enclosed by $C$:
$$\begin{aligned}
\oint_C f(z) \dd{z}
&= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
\end{aligned}$$
Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
equations, such that the integrands disappear and the final result is zero.
</div>
</div>
An interesting consequence is **Cauchy's integral formula**, which
states that the value of $f(z)$ at an arbitrary point $z_0$ is
determined by its values on an arbitrary contour $C$ around $z_0$:
$$\begin{aligned}
\boxed{
f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-int-formula"/>
<label for="proof-int-formula">Proof</label>
<div class="hidden">
<label for="proof-int-formula">Proof.</label>
Thanks to the integral theorem, we know that the shape and size
of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$,
such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then
we integrate by substitution:
$$\begin{aligned}
\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
&= \frac{1}{2 \pi i} \int_0^{2 \pi} f(z) \frac{i r e^{i \theta}}{r e^{i \theta}} \dd{\theta}
= \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
\end{aligned}$$
We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:
$$\begin{aligned}
\lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
&= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
= f(z_0)
\end{aligned}$$
</div>
</div>
Similarly, **Cauchy's differentiation formula**,
or **Cauchy's integral formula for derivatives**
gives all derivatives of a holomorphic function as follows,
and also guarantees their existence:
$$\begin{aligned}
\boxed{
f^{(n)}(z_0)
= \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z}
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-diff-formula"/>
<label for="proof-diff-formula">Proof</label>
<div class="hidden">
<label for="proof-diff-formula">Proof.</label>
By definition, the first derivative $f'(z)$ of a
holomorphic function exists and is:
$$\begin{aligned}
f'(z_0)
= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
\end{aligned}$$
We evaluate the numerator using Cauchy's integral theorem as follows:
$$\begin{aligned}
f'(z_0)
&= \lim_{z \to z_0} \frac{1}{z - z_0}
\bigg( \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \bigg)
\\
&= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0}
\oint_C \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta}
\\
&= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0}
\oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta}
\end{aligned}$$
This contour integral converges uniformly, so we may apply the limit on the inside:
$$\begin{aligned}
f'(z_0)
&= \frac{1}{2 \pi i} \oint_C \Big( \lim_{z \to z_0} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)} \Big) \dd{\zeta}
= \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta}
\end{aligned}$$
Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
this proof works inductively for all higher orders $n$.
</div>
</div>
|
