Conformal map onto unit disk

Question

I need to find a conformal map from $\mathbb{C} - (D^2 \cup [-2,-1] \cup [1, \infty))$. I've tried looking at rotations, exponential maps and mobius transforms, but have so far had no luck.

Answer 1

Let $U$ be the given domain. We want to map it conformally into the unit disk. I'm assuming $D^2$ means the unit disk.

First do $z \mapsto 1/z$. This maps $U$ into the domain $U_1$ which is just the unit circle, without the line segment from $0$ to $1$ and the line segment from $-1$ to $-1/2$. Call this $U_2$.

Then do $z \mapsto \frac{i(z+1)}{1-z}$. This maps $U_2$ into the upper half plane, without the line segment from $0$ to $i/3$, and the vertical ray from $i$ to $\infty$ along the imaginary axis. Call this $U_3$.

Then do $z \mapsto z^2$. This maps $U_3$ into the entire plane, without the horizontal rays from $-1/9$ to $\infty$ intersecting the positive real axis, and that from $\infty$ to $-1$ along the negative real axis. Call this $U_4$.

Then do $z \mapsto (9/8)z+1/8$, which maps $U_4$ into the domain without the rays from $0$ to $\infty$ along the positive real axis, and that from $\infty$ to $-1$ along the negative real axis. Call this $U_5$.

Then do $z \mapsto \sqrt{z}$ w.r.t. the branch of the argument from $0$ to $2 \pi$. This maps $U_5$ into the domain $U_6$ which is the upper half plane without the vertical ray from $i$ to $\infty$ along the imaginary axis.

Then do $z \mapsto -1/z$ which maps $U_6$ to the domain $U_7$ which is the upper half plane without the line segment from $0$ to $i$.

Then do $z \mapsto \sqrt{z^2+1}$ which sends $U_7$ into the upper half plane.

Then do $z \mapsto \frac{z-i}{z+i}$ which sends the upper half plane into the unit circle.

Finally, compose all of these maps.