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# proof of Casorati-Weierstrass theorem

Assume that $a$ is an essential singularity of $f$. Let $V\subset U$ be a punctured neighborhood of $a$, and let $\lambda\in\mathbb{C}$. We have to show that $\lambda$ is a limit point of $f(V)$. Suppose it is not, then there is an $\epsilon>0$ such that $|f(z)-\lambda|>\epsilon$ for all $z\in V$, and the function

$g:V\to\mathbb{C},z\mapsto\frac{1}{f(z)-\lambda}$ |

is bounded, since $|g(z)|=\frac{1}{|f(z)-\lambda|}<\epsilon^{{-1}}$ for all $z\in V$. According to Riemann’s removable singularity theorem, this implies that $a$ is a removable singularity of $g$, so that $g$ can be extended to a holomorphic function $\bar{g}:V\cup\{a\}\to\mathbb{C}$. Now

$f(z)=\frac{1}{\bar{g}(z)}-\lambda$ |

for $z\neq a$, and $a$ is either a removable singularity of $f$ (if $\bar{g}(z)\neq 0$) or a pole of order $n$ (if $\bar{g}$ has a zero of order $n$ at $a$). This contradicts our assumption that $a$ is an essential singularity, which means that $\lambda$ must be a limit point of $f(V)$. The argument holds for all $\lambda\in\mathbb{C}$, so $f(V)$ is dense in $\mathbb{C}$ for any punctured neighborhood $V$ of $a$.

To prove the converse, assume that $f(V)$ is dense in $\mathbb{C}$ for any punctured neighborhood $V$ of $a$. If $a$ is a removable singularity, then $f$ is bounded near $a$, and if $a$ is a pole, $f(z)\to\infty$ as $z\to a$. Either of these possibilities contradicts the assumption that the image of any punctured neighborhood of $a$ under $f$ is dense in $\mathbb{C}$, so $a$ must be an essential singularity of $f$.

## Mathematics Subject Classification

30D30*no label found*

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