Question about open and inverse mapping theorem Suppose $f$ is analytic on $D(z_{0}, R)$ \ {$z_{0}$}, and $z_{0}$ is a pole of $f$. Prove that for any $r \in (0, R)$, there is $M \in (0, \infty)$ such that $f(D(z_{0}, r))$ \ {$z_{0}$} $\supset$ {$|z| > M$}
I totally do not know how to start this problem. I know that the limit at the pole is infinity, but how can I show that all the points outside the circle {$|z| > M$} are in the image of $f$? And for the boundary $M$, I tried to apply the maximum modulus principle for harmoinc function, but it seems $f$ may not be defined at point $z_{0}$.
Any hint is appreciated, thanks a lot.
 A: Let $g(z) = \dfrac{1}{f(z)}$ (with $g(z_0) = 0$). Then $g$ is holomorphic on a neighbourhood $D_\varepsilon(z_0) = \{ |z-z_0| < \varepsilon \}$ of $z=z_0$, so by the open mapping theorem, $g$ maps $D_\varepsilon$ onto an open neighbourhood of $0$, $g(D_\varepsilon) \supset D_\delta(0)$ for some $\delta > 0$.
Hence $f$ maps $D_\varepsilon$ to a set containing $\{ |z| > 1/\delta \}$.
A: Since $z_0$ is a pole of $f$ we have $f(z) = {g(z) \over (z-z_0)^m}$ for some
$g$ analytic on $B(z_0,R)$ with $g(z_0) \neq 0$.
Let $\phi(z) = { (z-z_0)^m\over g(z)}$, then we see that $\phi$ is analytic on
some $B(z_0,\rho)$, with $0<\rho<R$ and $\phi(z_0) = 0$. 
Choose $0<r<R$, and let $r'=\min(\rho, r)$. Since $\phi$ is analytic, it is an
open map and hence $\phi(B(z_0,r'))$ is an open set containing $0$. Hence
$B(0,{1 \over M}) \subset \phi(B(z_0,r'))$ for $M$ sufficiently large.
In particular, for all $w$ such that $|w| > M$, there is some $z \in B(z_0,r')$
such that ${1 \over w} = {1 \over f(z)}$ (and so $z \neq z_0$, since ${ 1\over w} \neq 0$), and hence we see that
$\{ w | |w| > M \} \subset f(B(z_0,r') \setminus \{z_0\})$.
