---
title: "Holomorphic function"
sort_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" markdown="1">
<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" markdown="1">
<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" markdown="1">
<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>