kuniga.me > Docs > Holomorphic Functions Cheat Sheet
A complex function $f(z)$ is holomorphic at point $a$ if it’s complex differentiable at $a$. More formally the limit:
\[\lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} = f'(a)\]for $h \in \mathbb{C}$ exists. Without any qualifications, a holomorphic function is a function that is holomorphic at every point on its domain.
The following operations preserve holomorphism:
Let $f(z)$ and $g(z)$ be holomorphic functions in $\Omega$. Then the following functions are also holomorphic in $\Omega$:
Let $f(z)$ be a holomorphic function in $\Omega$.
If $f(z) \ne 0$ and $\Omega$ is simply connected, then we can choose a single-valued branch of $\log f(z)$ that is also holomorphic.
If $f(z) \ne 0$ but $\Omega$ is multiply connected, then we cannot choose one single-valued branch of $\log f(z)$.
If $f(z)$ is holomorphic, then $\exp(f(z))$ or $e^{f(z)}$ is also holomorphic.
If $f(z)$ is holomorphic in $\Omega$, then $f’(z)$ is also holomorphic in $\Omega$ [2].
If $f(z)$ is a holomorphic function in $\Omega$, there exists a holomorphic function $g(z)$ in $\Omega$ such that $g’(z) = f(z)$.
Holomorphic functions are conformal at a point $z$ if $f’(z) \ne 0$ [4].
Holomorphic functions are continuous.
Let $f(z)$ be a holomorphic function in the disk $\abs{z - a} \lt r$. Then, for $z$ in that circle it can be written as a Taylor series expansion around $a$:
\[f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (z - a)^n\]If we denote $c_n = f^{(n)}(a) / n!$, it looks more like a polynomial:
\[f(z) = \sum_{n=0}^{\infty} c_n (z - a)^n\]Let $f(z)$ be holomorphic and bounded by a finite $M$ in a region $\Omega$ (i.e. , $\abs{f(z)} \le M$ for all $z \in \Omega$). Let $C$ be a circle of radius $r$ centered in $a$ ($C$ is inside $\Omega$). Then:
\[\abs{f^{(n)}(a)} \le \frac{n! M}{r^{n}}\]In terms of the Taylor series coefficient:
\[\abs{c_n} \le \frac{M}{r^n}\]If $f(z)$ is holomorphic and bounded on the whole plane, then it’s a constant function [2].