Apparently, the following statement is true:
- "Let $D\subseteq \mathbb{C}$ be open and connected and $f:D\setminus \{a\}\longrightarrow \mathbb{C}$ holomorphic with a pole of arbitrary order at $a\in D$. For any $\epsilon > 0$ with $B_\epsilon(a)\setminus\{a\} \subseteq D$, there exists $r > 0$ so that $\{z \in \mathbb{C}: |z| > r\} \subseteq f(B_\epsilon(a)\setminus\{a\})$."
So far, I have been unsuccessful in proving this. I know that $f(B_\epsilon(a)\setminus\{a\})$ must be open and connected (open mapping theorem), as well as that for any $r > 0$ there exists an $x \in B_\epsilon(a)$ so that $f(x) > r$ (because $\lim_{z\rightarrow a}|f(z)| = \infty)$, but I don't see how this would imply the statement in question. Any help would be appreciated.