I'm struggling with the following problem. Any help would be a really welcomed.
Let $f$ a meromorphic function on $\mathbb C$ s.t $g(w)= f(\frac{1}{w})$ is meromorphic for $w=0$. We say that $f$ is meromorphic over $\hat{ \mathbb C}$.
I proved that $f$ can be extended to an analytic (i.e continuous) function $\hat{f}:\hat{ \mathbb C}\rightarrow \hat{ \mathbb C}$ but I have some trouble proving:
- $\hat{f}$ is onto if $f$ is not a constant.
- If $\hat{f}$ is bijective, then $f$ is a Mobius function.
For 1) I tried using the compactness of $\hat{ \mathbb C}$ and the open mapping theorem but I failed.
For 2) I know that $\hat{f}$ is bijective i.e a holomorphic automorphism of $\hat{ \mathbb C}$ and that if $f$ is Mobius then $f(z)= \frac{ax+b}{cz+d},\forall a,b,c,d\in \mathbb C$ s.t $ac-bd \neq 0$ , but I don't know how I can proceed.
Thanks in advance for your help.
If $f$ is not constant, it is open. Since its domain ($\hat{\mathbb{C}}$) is compact, $f$ is also closed, so $f(\hat{\mathbb{C}})$ is clopen and not empty, a contradiction since $\hat{\mathbb{C}}$ is connected: a connected set $A$ cannot have a non empty clopen subset $B$, since otherwise $A=B\sqcup A\backslash B$ with both $B, A\backslash B$ open set.
Since $f$ is meromorphic at $\infty$, it must be a rational function (every non rational function has an essential singularity at $\infty$, can you prove it?). Thus $f=\frac{P(z)}{Q(z)}$. Let us suppose that $P,Q$ are coprime. Since it is bijective, we can have only one point sent to $\infty$, i.e. one and only one pole, which implies $\text{deg}(Q)\le1$. Similarly, $\#f^{-1}(0)=1$, so $\text{deg}(P)\le 1$. We can now write $f(z)=\frac{az+b}{cz+d}$, and it is straightforward to prove that since $f$ is bijective $ad-bc\neq 0$, which gives us the claim.