Suppose $f$ is analytic on $D(z_0;R)\setminus\{z_0\}$, and $z_0$ is a pole of $f$. Prove that for any $r\in(0,R)$, there is $M\in(0,\infty)$ such that $f(D(z_0;r)\setminus\{z_0\})\supset\{z\in\mathbb{C}:|z|>M\}$.
My thoughts:
Consider $f(D(z_0;r)\setminus\{z_0\})$. What can we say about this set? We know that $f$ is holomorphic on $D(z_0;r)\setminus\{z_0\}$. Also, $f(D(z_0;r)\setminus\{z_0\})$ would be an open set if $D(z_0;r)\setminus\{z_0\}$ were open, according to the Open Mapping Theorem (but alas, I'm not sure if $D(z_0;r)\setminus\{z_0\}$ is open). Then, we could use the definition of open to find discs centered around any point in the set $f(D(z_0;r)\setminus\{z_0\})$ which stay in the set.
Am I on to something? I know I am missing a lot of details. Thanks for your help!
Since $z_0$ is a pole of $f$, $f(z) = \frac{1}{h(z)}$, where $h$ is analytic on an open disc $D(z_0;\delta)$ centered at $z_0$ and $h(z_0) = 0$. Let $r \in (0,R)$ and set $\rho = \min\{r,\delta\}$. By the open mapping theorem, $h(D(z_0;\rho))$ is open. Since $0 = h(z_0)\in h(D(z_0;\rho))$, there exists a positive number $k$ such that $h(D(z_0;\rho)) \supset \{z:|z| < k\}$. Let $M = \frac{1}{k}$. If $|z| > M$, then $|\frac{1}{z}| < k$, which implies $\frac{1}{z} = h(u) = \frac{1}{f(u)}$ for some $u\in D(z_0;\rho)$. Since $u\neq z_0$ and $z = f(u)$, $$z \in f(D(z_0;\rho)\setminus\{z_0\})\subset f(D(z_0;r)\setminus\{z_0\}.$$ As $z$ was arbitrary, $f(D(z_0;r)\setminus\{z_0\}) \supset \{z:|z| > M\}$.