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 $f$ .