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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.

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I have changed \diagdown to \setminus – user9413 Jul 6 '11 at 15:28
Have you considered the function $1/f$ near $a$? Depending on the definitions used in your text, you may immediately have that $1/f$ is a holomorphic (and hence continuous) function near $a$, that extends continuously to $1/f(a) = 0$. – Willie Wong Jul 6 '11 at 15:40
Do you know about the Riemann sphere? This is basically saying that the function is somehow "continuous" through $\infty$... which can be made precise by studying the behavior of $1/f$. This is especially clear if you look at Laurent series. – Aaron Mazel-Gee Jul 6 '11 at 15:41
@Aaron: It is as much about openness as continuity. – Jonas Meyer Jul 6 '11 at 15:45
@Jonas: That's true, thanks. Of course, all nonconstant holomorphic maps are open, so I guess it could depend on what we're taking as known.. – Aaron Mazel-Gee Jul 6 '11 at 20:03
up vote 8 down vote accepted

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 $r=\frac{1}{\delta}$.

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