---
title: "Holomorphic function"
firstLetter: "H"
publishDate: 2021-02-25
categories:
- Mathematics

date: 2021-02-25T14:40:45+01:00
draft: false
markup: pandoc
---

# Holomorphic function

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:

$$\begin{aligned}
    \oint_C f(z) \:dz
    &= \oint_C (u + i v) \dd{(x + i y)}
    = \oint_C (u + i v) \:(\dd{x} + i \dd{y})
    \\
    &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
\end{aligned}$$

Using Green's theorem, we integrate over the area $A$ enclosed by $C$:

$$\begin{aligned}
    \oint_C f(z) \:dz
    &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
\end{aligned}$$

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>


## Residue theorem

A function $f(z)$ is **meromorphic** if it is holomorphic except in
a finite number of **simple poles**, which are points $z_p$ where
$f(z_p)$ diverges, but where the product $(z - z_p) f(z)$ is non-zero and
still holomorphic close to $z_p$.

The **residue** $R_p$ of a simple pole $z_p$ is defined as follows, and
represents the rate at which $f(z)$ diverges close to $z_p$:

$$\begin{aligned}
    \boxed{
        R_p = \lim_{z \to z_p} (z - z_p) f(z)
    }
\end{aligned}$$

**Cauchy's residue theorem** generalizes Cauchy's integral theorem
to meromorphic functions, and states that the integral of a contour $C$
depends on the simple poles $p$ it encloses:

$$\begin{aligned}
    \boxed{
        \oint_C f(z) \dd{z} = i 2 \pi \sum_{p} R_p
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-res-theorem"/>
<label for="proof-res-theorem">Proof</label>
<div class="hidden">
<label for="proof-res-theorem">Proof.</label>
From the definition of a meromorphic function,
we know that we can decompose $f(z)$ like so,
where $h(z)$ is holomorphic and $p$ are all its poles:

$$\begin{aligned}
    f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p}
\end{aligned}$$

We integrate this over a contour $C$ which contains all poles, and apply
both Cauchy's integral theorem and Cauchy's integral formula to get:

$$\begin{aligned}
    \oint_C f(z) \dd{z}
    &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
    = \sum_{p} R_p \: 2 \pi i
\end{aligned}$$
</div>
</div>

This theorem might not seem very useful,
but in fact, thanks to some clever mathematical magic,
it allows us to evaluate many integrals along the real axis,
most notably [Fourier transforms](/know/concept/fourier-transform/).
It can also be used to derive the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations).