Prove: if $f \in Hol(\mathbb{C}\setminus \{0\}),$ if $f$ is one to one and onto, then $f = az$ or $f = \frac{a}{z}$

Question

Prove: if $f \in Hol(\mathbb{C}\setminus \{0\}),$ if $f$ is one to one and onto, then $f = az$ or $f = \frac{a}{z}$

I know that in order to solve this I have to examine the singularities of $f$ at the points $0$ and $\infty$. I've proved that these cases can't exist:

1. both of them are removable singularities
2. either of them is an essential singularity

I'm having trouble disproving the case in which both of the points are poles. Also, I can't think of the case in which one of the singularities is a pole and the other is removable - why does the pole have to be simple.

I think perhaps the way I've approached the problem is at fault. It seem's like a constructive proof would be simpler.

Any help would be appreciated!

Answer 1

Note that meromorphic functions on the Riemann sphere are rational. Since you've dealt with the case where either singularity is essential, this means that you only need to look at rational functions whose only poles are at $0$ and $\infty$. Fortunately, the supply of those is relatively limited...