
conformal selfmap
Show that any conformal selfmap of the upper halfplane has the form
$\displaystyle f(z)=\frac{az+b}{cz+d}$, $\displaystyle \text{Im}(z)>0$,
where $\displaystyle a, b, c, d$ are real numbers satisfying $\displaystyle adbc=1$. When do two such coefficient choices for $\displaystyle a, b, c, d$ determine the same conformal selfmap of the upper halfplane?
In this section we have covered Pick's Lemma and we have covered that fractional linear transformations have the form $\displaystyle w=f(z)=\frac{az+b}{cz+b}$. However, I do not see how conformal selfmaps in the upper halfplane have this form too. I need some hints on how to start this one. Thanks in advance.

First, show that such a function must have only one simple pole in the extended complex plane. Then subtract the principal part, so that you are left with an analytic function with no pole in the extended plane; by Liouville's theorem, such a function must be a constant. So $\displaystyle f(z)=\frac{1}{za} + C$ is of the form you describe.