Behaviour of a holomorphic function near a pole Apparently, the following statement is true:

  
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*"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.

Define $g$ on a punctured neighborhood of $a$ by $g(z)=\frac{1}{f(z)}$.  Then $\displaystyle{\lim_{z\to a}g(z)=0}$, so the singularity of $g$ at $a$ is removable, and defining $g(a)=0$ gives an analytic function on a neighborhood of $a$.  By the open mapping theorem, for each neighborhood $U$ of $a$ in the domain of $g$, there exists $\delta>0$ such that $\{z\in\mathbb{C}:|z|\lt \delta\}\subseteq g(U)$.  Now let $r=\frac{1}{\delta}$.
