# Conformal transformation of complement of disk in upper half plane

Let $U$ be the complement in the half-plane $\operatorname{Im} z > 0$ of a disk of radius $a<1$ centered at $i$. I am looking for a conformal transformation that maps $U$ onto an annulus. Since $U$ is not simply connected, I am not certain if such a transformation exists. Does such a transformation exist, and can it be written in a simple form?

Ultimately my goal is to solve the Laplace equation in $U$ with Dirichlet boundary conditions.

• $U$ is simply connected. You could transform it into a disc, but not into an annulus. – Pp.. Jan 28 '15 at 13:56
• @Pp The radius of the disk is $a<1$, so the disk and the real line do not 'touch'. Therefore $U$ is not simply connected. – user111187 Jan 28 '15 at 14:03
• You are right. Didn't read where the center was. Then the idea is: Send the upper half-plane to the disc. Then $U$ is sent to the region between two not-necessarily concentric circles. Now, prove that for every two not-touching circles you can find a transformation that makes them concentric. – Pp.. Jan 28 '15 at 14:03
• @Pp Ah, because Möbius transformations send circles to circles! That is smart. I think I can figure it out with this. – user111187 Jan 28 '15 at 14:12
• Are you successful? – ts375_zk26 Jan 28 '15 at 21:01

## 1 Answer

First we note that $$w=f(z)=\frac{z-\alpha i}{z+\alpha i}$$ with $\alpha >0$ maps the upper half-plane to the unit disc and that $f(z)$ sends the circle of radius $a<1$ centered at $i$ to a circle $C=f(\{|z-i|=a\})$, whose center and radius are unknown.
But we know by the symmetry that the center of $C$ lies on the real axis. Also this is easily seen from the fact that $f(yi)$ is real.
If we choose $\alpha$ so that $f((1+a)i)=-f((1-a)i)$,then $C$ and $|w|=1$ are concentric by the symmetry. A bit calculation leads to $\alpha =\sqrt{1-a^2}$. Note that the radius of $C$ is determined automatically (we can not specify it).