Conformal map onto upper-half plane.

Question

What is a conformal map of the set$$\left\{z \in \mathbb{C} : |z| < 1 \text{ and }\left|z - {1\over2}\right| > {1\over2}\right\}$$onto the upper half plane $\{z \in \mathbb{C} : \text{Im}(z) > 0\}$? I probably first want to find a conformal transformation onto a horizontal strip in $\mathbb{C}$, but then what?

Answer 1

Hint: you can transform an horizontal strip into an angle with vertex at the origin by means of the exponential function.

Answer 2

Call the domain above $D$. We first find a fractional linear transformation which maps $1$ to $\infty$, so the two boundary circles map to parallel lines in $\mathbb{C}$. The function $1/(z-1)$ maps $D$ to the vertical strip$$\left\{z : -1 < \text{Re}(z) < -{1\over2}\right\}.$$Multiplying by $2\pi i$ and adding $2\pi i$ gives the function$${{2\pi i}\over{z - 1}} + 2\pi i = {{2\pi iz}\over{z-1}},$$which maps $D$ to the horizontal strip$$\{z : 0 < \text{Im}(z) < \pi\}.$$Finally, we can apply the exponential map $e^z$ to map this to the upper half-plane. Thus, the function$$f(z) = e^{{2\pi iz}\over{z-1}}$$does what we need.

Answer 3

Hint: Any point in the upper half plane an be written as $re^{i\phi}=e^{\ln r+i\phi}$ with $0<\phi<\pi$ and $r>0$ (i.e., $-\infty<\ln r<\infty$).