Map the complement of the arc $|z|=1$, $Imz\geq0$, on the outside of the unit circle so that points at $\infty$ correspond to each other. Question: Map the complement of the arc $|z|=1$, $\operatorname{Im}z\geq0$, on the outside of the unit circle so that points at $\infty$ correspond to each other.
My attempt: Let $\Omega$ be the region we start with above.
$z_1\mapsto\frac{z_1+1}{z_1-1}$ takes $\Omega$ to the upper half plane.
$z_2\mapsto\sqrt{z_2}$ takes the upper half plane to the right upper half plane.
$z_3\mapsto\frac{z_3-1}{z_3+1}$ takes the right upper half plane to the upper unit circle.
$z_4\mapsto z_4^2%$ takes the upper unit circle to the unit circle.
$z_5\mapsto\frac{1}{z_5}$ takes the unit circle to the outside of the unit circle.
Finally, composing them, $z_5z_4z_3z_2z_1z=\frac{1}{\Bigg(\frac{\sqrt{\frac{z+1}{z-1}}-1}{\sqrt{\frac{z+1}{z-1}+1}}\Bigg)^2}$
I was wondering if I am correct, or if there is a glaring mistake somewhere.  Moreover, how do I confirm that the points at $\infty$ correspond to each other?  Do I need to consider the cross product?  Any help is greatly appreciated!  Thank you.
I did notice that this problem is asked here:
Conformal mapping of nonsimply connected domains
but I am just wondering whether or not my map would preserve the points at $\infty$, or if there is something more...
 A: Let $\Omega$ be the complex plane with the arc $\{e^{it}:t \in [0,\pi]\}$ removed.
The following steps, when composed, map $\Omega$ onto $\mathbb{C}\setminus\overline{\mathbb{D}}$, as desired.


*The starting point is $\Omega$.


*Apply $T_1: z\mapsto \frac{az+b}{cz+d}=\frac{z+1}{z-1}.$


*Apply $T_2: z\mapsto \sqrt{z}=(re^{i\theta})^{\frac{1}{2}}$,   $\quad\{-\frac{\pi}{2}<\theta<\frac{3\pi}{2}\}.$


*Apply $T_3:z \mapsto e^{i\pi/4}z$.


*Apply $T_4: z\mapsto \frac{az+b}{cz+d}=\frac{z-e^{i\pi/4}}{z-e^{-i\pi/4}}.$


*Apply $T_5: z\mapsto \frac{az+b}{cz+d}=\frac{1}{z}.$
The answer given by Hagen von Eitzen during the year '14 is not very different. He began like this:


*The starting point is $\Omega$.


*Apply $H_1: z\mapsto \frac{az+b}{cz+d}=\frac{1}{z+1}.$


*Apply $H_2:z \mapsto z-\frac{1}{2}$.


*Apply a branch of square root...
from there, the finish is similar to the one above with the $T_j.$
You should double check that the point at infinity returns back home after its journey around $\overline{\mathbb{C}}.$
