Let $U \subset \mathbb{C}$ be open, let $a \in U$, and let $f: U \setminus \{a\} \to \mathbb{C}$ be holomorphic. For $c \in \mathbb{C}$ and $r > 0$, write $B(c,r) := \{z \in \mathbb{C} : |z-c| < r\}$, and write $\mathbb{N} := \{1,2,3,\dots\}$. Assume that $a$ is called a pole of $f$ precisely when there exist $r > 0$ with $B(a,r) \subset U$, an integer $m \in \mathbb{N}$, and a holomorphic map $g: B(a,r) \to \mathbb{C}$ such that $g(a) \neq 0$ and

\begin{align*} f(z) = \frac{g(z)}{(z-a)^m} \end{align*}

for every $z \in B(a,r) \setminus \{a\}$. Then $a$ is a pole of $f$ if and only if, for every $M > 0$, there exists $\delta > 0$ such that

\begin{align*} 0 < |z-a| < \delta \text{ and } z \in U \implies |f(z)| > M. \end{align*}

Equivalently,

\begin{align*} \lim_{z \to a} |f(z)| = \infty \end{align*}

in the punctured-neighborhood sense inside $U$.