I'm working on a problem that was a previous complex qualifying exam at my university. I believe I have a solution, but I'm not entirely confident in it. The problem is as follows:
Find a one-to-one conformal map of the region $\Omega=\{z\in\mathbb{C}\,|\,|z|<2\text{ and }|z-1|>1\}$ onto the upper half plane.
Here's my attempt:
The region $\Omega$ can be mapped to the vertical strip $\{z\in\mathbb{C}\,|\, -1/2<\text{Re }(z)<-1/4\}$ by the map $z\mapsto\frac{1}{z-2}$. Then $z\mapsto\left(\frac{4\pi}{z-2}+2\pi\right)i$ maps $\Omega$ to the horizontal strip $\{z\in\mathbb{C}\,|\,0<\text{Im }(z)<\pi\}$. Lastly $$ z\mapsto \exp\left(\frac{4\pi}{z-2}i+2\pi i\right) $$ or $$ z\mapsto\exp\left(\frac{4\pi}{z-2}i\right) $$ should map to the upper half plane.
Is this correct? Is there any easier way to see this? Any input would be greatly appreciated, thanks in advance.
This is correct, and there is no easier approach. Any time you have a domain with two mutually tangent circles in its boundary, you are probably going to send the point of tangency to $\infty$, to turn the circles into parallel lines. And the subsequent step of mapping an infinite strip onto halfplane with exponential functions is also standard.