Casorati-Weierstrass theorem

First, assume that ${\displaystyle z_{0}}$ is an essential singularity of ${\displaystyle f}$, and suppose there exists an ${\displaystyle r>0}$ such that${\displaystyle f(B_{r}(z_{0})\setminus \{z_{0}\})}$ is not dense in ${\displaystyle \mathbb {C} }$. Then there exists an ${\displaystyle \epsilon >0}$ and a ${\displaystyle w_{0}\in \mathbb {C} }$ such that ${\displaystyle B_{\epsilon }(w_{0})}$ and ${\displaystyle f(B_{r}(z_{0})\setminus \ {z_{0}})}$ are disjoint. Consider ${\displaystyle B_{r}(z_{0})\setminus \{z_{0}\}}$ the function. ${\displaystyle g(z):={\frac {1}{f(z)-w_{0}}}}$.Let ${\displaystyle r}$ be chosen so that ${\displaystyle z_{o}}$ is the only ${\displaystyle w_{0}}$-pole in ${\displaystyle f(B_{r}(z_{0}))}$. This is possible by the Identity Theorem for non-constant holomorphic functions. Since ${\displaystyle f}$ is not constant (as it has an essential singularity), it is holomorphic and bounded by ${\displaystyle {\frac {1}{\epsilon }}}$. By the Riemann Removability Theorem, ${\displaystyle g}$ is therefore holomorphically extendable to all of ${\displaystyle B_{r}(z_{0})}$. Since ${\displaystyle g\neq 0}$ there exists an ${\displaystyle m\geq 0}$ and a holomorphic function ${\displaystyle g_{0}\colon B_{r}(z_{0})\to \mathbb {C} }$ with ${\displaystyle g_{0}(z_{0})\neq 0}$, such that

${\displaystyle g(z)=(z-z_{0})^{m}g_{0}(z),\qquad |z-z_{0}|<r.}$

It follows that

${\displaystyle f(z)=w_{0}+{\frac {1}{(z-z_{0})^{m}g_{0}(z)}}}$

and thus

${\displaystyle f(z)(z-z_{0})^{m}=w_{0}(z-z_{0})^{m}+{\frac {1}{g_{0}(z)}}}$

Since ${\displaystyle g_{0}(z_{0})\neq 0}$,is ${\displaystyle {\frac {1}{g}}}$ is holomorphic in a neighborhood of ${\displaystyle z_{0}}$. Therefore, ${\displaystyle f\cdot (\cdot -z_{0})^{m}}$ is holomorphic in a neighborhood of ${\displaystyle z_{0}}$, meaning that ${\displaystyle f}$ has at most a pole of order ${\displaystyle m}$ at ${\displaystyle z_{0}}$, which leads to a contradiction.Conversely. let ${\displaystyle z_{0}}$ be a removable singularity or a pole of ${\displaystyle f}$. Is ${\displaystyle z_{0}}$ is a removable singularity, there exists a neighborhood ${\displaystyle U}$ of ${\displaystyle z_{0}}$,where ${\displaystyle f}$ is bounded, say ${\displaystyle |f(z)|\leq M}$ for ${\displaystyle z\in U\setminus \{z_{0}\}}$.Then it follows that

${\displaystyle {\overline {f{\bigl (}U\setminus \{z_{0}\})}}\subseteq {\bar {B}}_{M}(0)\neq \mathbb {C} .}$

If ${\displaystyle z_{0}}$ is a pole of order ${\displaystyle m}$ for ${\displaystyle f}$, there exists a neighborhood ${\displaystyle U}$ of ${\displaystyle z_{0}}$ and a holomorphic function ${\displaystyle g\colon U\to \mathbb {C} }$ with ${\displaystyle g(z_{0})\neq 0}$ and ${\displaystyle f(z)=g(z)(z-z_{0})^{-m}}$. Choose a neighborhood ${\displaystyle \epsilon >0}$ such that ${\displaystyle |g(z)|\geq {\frac {1}{2}}|g(z_{0})|}$ for ${\displaystyle |z-z_{0}|<\epsilon }$. Then it follows that

${\displaystyle |f(z)|=|g(z)||z-z_{0}|^{-m}\geq {\frac {1}{2}}|g(z_{0})|\cdot \epsilon ^{-m},\quad 0<|z-z_{0}|<\epsilon }$

Thus, ${\displaystyle 0\not \in {\overline {f(B_{\epsilon }(z_{0})\setminus \{z_{0}\})}}}$ and this proves the claim.