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 $\lambda \in ℂ$.We have to show that $\lambda$ is a limit point of $f(V)$. Suppose itis not, then there is an $ϵ>0$ such that$|f(z)-\lambda |>ϵ$ for all $z\in V$, and the function

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

is bounded, since for all $z\in V$. According to Riemann’s removable singularitytheorem, this implies that $a$ is a removable singularity of $g$, sothat $g$ can be extended to a holomorphic function $\overline{g}:V\cup \{a\}\to ℂ$. Now

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

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

To prove the converse, assume that $f(V)$ is dense in $ℂ$ forany punctured neighborhood $V$ of $a$. If $a$ is a removablesingularity, then $f$ is bounded near $a$, and if $a$ is a pole,$f(z)\to \mathrm{\infty }$ as $z\to a$. Either of these possibilitiescontradicts the assumption that the image of any puncturedneighborhood of $a$ under $f$ is dense in $ℂ$, so $a$ must bean essential singularity of $f$.