# conformal self-map

• Feb 10th 2010, 11:12 AM
eskimo343
conformal self-map
Show that any conformal self-map of the upper half-plane 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 $\displaystyle ad-bc=1$. When do two such coefficient choices for $\displaystyle a, b, c, d$ determine the same conformal self-map of the upper half-plane?

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 self-maps in the upper half-plane have this form too. I need some hints on how to start this one. Thanks in advance.
• Feb 10th 2010, 11:58 AM
Bruno J.
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 $\displaystyle f(z)=\frac{1}{z-a} + C$ is of the form you describe.