Image of a neighborhood of a pole

Question

Let $f$ be a meromorphic function and $a$ a pole of $f$. Let $U$ be a neighborhood of $a$. Is it true that $f(U)$ contains $\{z| |z|>r\}$ for some $r$? All I know is that $\lim\limits_{z\rightarrow a} |f(z)|=\infty$.

Thanks for your help.

Answer 1

Yes, it is true.

Let $n$ be the order of your pole. Then, for some neighborhood $U$ of $a$,$$z\in U\setminus\{a\}\implies f(z)=(z-a)^{-n}g(z)$$for some analytic function $g$ such that $g(a)\neq0$. Since, in $U\setminus\{a\}$, you always have $f(z)\ne0$, you can define (again, in $U\setminus\{a\}$) $h(z)=\frac1{f(z)}$, and then$$\lim_{z\to a}h(z)=\lim_{z\to a}\frac{(z-a)^n}{g(z)}=0,$$and therefore $0$ is a removable singularity of $h$ and, if you extend $h$ to $U$ defining $h(a)=0$, then you still have an analytic function. By the open mapping theorem, there is some $\varepsilon>0$ such that $h(U)\supset\{z\in\Bbb C\mid|z|<\varepsilon\}$. But then $f(U)\supset\left\{z\in\Bbb C\,\middle|\,|z|>\frac1\varepsilon\right\}$.