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.
Addition, subtraction and multiplication preserve holomorphism. Division preserves holomorphism as long as the divisor is non-zero [1].
Differetiation preserves holomorphism. In fact holomorphic functions are infinitely differentiable [2].
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:
\[\abs{c_n} \le \frac{M}{r^n}\]