Complex Analysis/Local theory of holomorphic functions

Proposition (holomorphic function has Taylor expansion convergent on its domain):

Let $U\subseteq \mathbb {C}$ be open, and let $f:U\to \mathbb {C}$ be holomorphic. Let $z_{0}\in U$. Then there exists coefficients $a_{0},a_{1},\ldots ,a_{n},\ldots$ of a Taylor series such that whenever ${\overline {B_{r}(z_{0})}}\subseteq \mathbb {C}$ is a ball of radius $r>0$ that is contained within $U$, then the Taylor series

$\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$

converges absolutely on ${\overline {B_{r}(z_{0})}}$ and equals $f(z)$ there.

Proof: By Cauchy's formula, we have

$f(z)=\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z}}dw$.

Adding a zero, we may rewrite this as

$f(z)=\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z_{0}-z-z_{0}}}dw$.

Computing further, and using the convergence of the geometric series for arguments of modulus $<1$,

${\begin{aligned}f(z)&=\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z_{0}-z-z_{0}}}dw\\&=\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{1-{\frac {z-z_{0}}{w-z_{0}}}}}{\frac {1}{w-z_{0}}}dw\\&=\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z_{0}}}\sum _{n=0}^{\infty }\left({\frac {z-z_{0}}{w-z_{0}}}\right)^{n}dw\\&=\sum _{n=0}^{\infty }\underbrace {\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}dw} _{=:a_{n}}(z-z_{0})^{n};\end{aligned}}$

here, interchanging differentiation and integration is justified by the absolute convergence of the former expression. $\Box$

Theorem (identiy theorem):

Let $U\subseteq \mathbb {C}$ be open, and let $f,g:U\to \mathbb {C}$ be two holomorphic functions defined on $U$. Suppose that the point $z_{0}\in U$ is an accumulation point of the set $\{z\in U|f(z)=g(z)\}$. Then $f(z)=g(z)$ for all $z$ in the connected component of $z_{0}$ of $U$.

Proof: We develop both $\Box$

Theorem (Riemann's theorem on removable singularities):

Let $f:U\to \mathbb {C}$ be a function which is holomorphic on $U\setminus \{z_{0}\}$ for a certain $z_{0}\in U$, and furthermore bounded in a ball (or rather disk) about $z_{0}$, say $B_{r}(z_{0})\subseteq U$, where $r>0$ is just the radius of that small ball. Then we will find a value $w_{0}$ such that if we continue $f$ to $w_{0}$ at the point $z_{0}$, the result will be holomorphic.

Proof: We define the new function

$g:U\to \mathbb {C} ,h(z)={\begin{cases}(z-z_{0})^{2}f(z)&z\neq z_{0}\\0&z=z_{0}\end{cases}}$

and claim that it's holomorphic on $U$. It will be complex differentiable in $U\setminus \{z_{0}\}$ "by the product rule" (which is supposed here to include the statement that the product of complex differentiable functions is complex differentiable), and in $z_{0}$ we have

$\lim _{g\to 0}{\frac {g(z_{0}+h)-g(z_{0})}{h}}={\frac {h^{2}(f(z_{0}+h)-f(z))}{h}}=0$

since $h$ is bounded.

Now we develop $h$ into a Taylor series at $z_{0}$. But the first two coefficients of it will be zero, whence $h$ will be divisible by $(z-z_{0})^{2}$ to obtain a power series, and this power series will define a holomorphic function about $z_{0}$. But by the definition of $h$, this series coincides with $f$ everywhere except $z_{0}$, whence if $f$ is continued at $z_{0}$ by the constant term of the series, the result will be holomorphic and thus the desired continuation. $\Box$

smoothness, integrability