# Casorati-Weierstrass theorem

The Casorati-Weierstrass theorem is a statement about the behavior of Holomorphic function in the vicinity of Isolated singularity. It essentially states that in every neighborhood of an essential singularity, every complex number can be arbitrarily closely approximated by the values of the function. It is a significantly easier-to-prove weakening of the great Picard theorem, which states that in every neighborhood of an essential singularity, every complex number (except possibly one) occurs infinitely often as a value.

## Statement

Let $G\subseteq \mathbb {C}$ be open, and $z_{0}\in G$. Let $f\colon G\setminus \{z_{0}\}\to \mathbb {C}$ be a Holomorphic function. Then, $f$ has an Isolated singularity at $z_{0}$ if and only if for every neighborhood $U\subseteq G$ of $z_{0}$: ${\overline {f(U\setminus {z_{0}})}}=\mathbb {C}$.

### Proof

First, assume that $z_{0}$ is an essential singularity of $f$, and suppose there exists an $r>0$ such that$f(B_{r}(z_{0})\setminus \{z_{0}\})$ is not dense in $\mathbb {C}$. Then there exists an $\epsilon >0$ and a $w_{0}\in \mathbb {C}$ such that $B_{\epsilon }(w_{0})$ and $f(B_{r}(z_{0})\setminus \ {z_{0}})$ are disjoint. Consider $B_{r}(z_{0})\setminus \{z_{0}\}$ the function. $g(z):={\frac {1}{f(z)-w_{0}}}$.Let $r$ be chosen so that $z_{o}$ is the only $w_{0}$-pole in $f(B_{r}(z_{0}))$. This is possible by the Identity Theorem for non-constant holomorphic functions. Since $f$ is not constant (as it has an essential singularity), it is holomorphic and bounded by ${\frac {1}{\epsilon }}$. By the Riemann Removability Theorem, $g$ is therefore holomorphically extendable to all of $B_{r}(z_{0})$. Since $g\neq 0$ there exists an $m\geq 0$ and a holomorphic function $g_{0}\colon B_{r}(z_{0})\to \mathbb {C}$ with $g_{0}(z_{0})\neq 0$, such that
$g(z)=(z-z_{0})^{m}g_{0}(z),\qquad |z-z_{0}|<r.$
It follows that
$f(z)=w_{0}+{\frac {1}{(z-z_{0})^{m}g_{0}(z)}}$
and thus
$f(z)(z-z_{0})^{m}=w_{0}(z-z_{0})^{m}+{\frac {1}{g_{0}(z)}}$
Since $g_{0}(z_{0})\neq 0$,is ${\frac {1}{g}}$ is holomorphic in a neighborhood of $z_{0}$. Therefore, $f\cdot (\cdot -z_{0})^{m}$ is holomorphic in a neighborhood of $z_{0}$, meaning that $f$ has at most a pole of order $m$ at $z_{0}$, which leads to a contradiction.Conversely. let $z_{0}$ be a removable singularity or a pole of $f$. Is $z_{0}$ is a removable singularity, there exists a neighborhood $U$ of $z_{0}$,where $f$ is bounded, say $|f(z)|\leq M$ for $z\in U\setminus \{z_{0}\}$.Then it follows that
${\overline {f{\biglU\setminus \{z_{0}\})}}\subseteq {\bar {B}}_{M}(0)\neq \mathbb {C} .$
If $z_{0}$ is a pole of order $m$ for $f$, there exists a neighborhood $U$ of $z_{0}$ and a holomorphic function $g\colon U\to \mathbb {C}$ with $g(z_{0})\neq 0$ and $f(z)=g(z)(z-z_{0})^{-m}$. Choose a neighborhood $\epsilon >0$ such that $|g(z)|\geq {\frac {1}{2}}|g(z_{0})|$ for $|z-z_{0}|<\epsilon$. Then it follows that
$|f(z)|=|g(z)||z-z_{0}|^{-m}\geq {\frac {1}{2}}|g(z_{0})|\cdot \epsilon ^{-m},\quad 0<|z-z_{0}|<\epsilon$
Thus, $0\not \in {\overline {f(B_{\epsilon }(z_{0})\setminus \{z_{0}\})}}$ and this proves the claim.

## see also

* Complex Analysis
* Identity Theorem

## Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:
* Source: Satz von Casorati-Weierstraß - URL:
https://de.wikiversity.org/wiki/Satz_von_Casorati-Weierstraß
* Date: 1/2/2025