Using Picard's Great theorem to show f isn't injective

Question

I've always seen it said as a simple or trivial result of Picard's great theorem that if a function f is analytic in a neighborhood near an essential singularity then f is not injective on this neighborhood, but have to show it and can't quite figure out how. It's probably simple but does anyone have a good way to show this is true?

Answer 1

Suppose otherwise. That is, let $V$ be a neighborhhod of $a$ and assume that $f$ has an essential singularity at $a$ and that $f|_{V\setminus\{a\}}$ is injective. Let $z_1$ and $z_2$ be two distinct points from $V\setminus\{a\}$. Then $V\setminus\{z_1,z_2\}$ is another neighborhood of $a$ and $f(z_1),f(z_2)\notin f\left(V\setminus\{a,z_1,z_2\}\right)$. So, there are two complex numbers which are not in the image of $f|_{V\setminus\{a,z_1,z_2\}}$, which is impossible, by Picard's great theorem.