Find a conformal from the upper-half plane with semi-disk removed to the unit disk

How do I find a conformal map from $W=\{Im(z)>0,|z|>1\}$, that is, the upper-half plane with semi-disk removed, onto the unit disk.

My first thought involves applying $\phi(z)=\frac{1}{z^2}$ or maybe the usual map from the upper-half plane to the unit disk $\psi(z)=\frac{z-i}{z+i}$. But I'm not really sure how to proceed. Any hints would be appreciated.

If you map $z=1, -1$ to $w=0, \infty$ then all boundary parts of $W$ become straight line segments, and it is not difficult to see that $T(z) = (z-1)/(z+1)$ maps $W$ to the first (upper right) quadrant.

Then continue with a mapping to the upper half plane and from there to the unit disk.

• @mantra: $T$ is a Möbiustransformation. Möbiustransformations are bijective mappings of the (extended) complex plane onto itself, and map lines or circles onto lines or circles. The idea is to transform the semicircle into a line. – Martin R Sep 16 '16 at 12:31
• @mantra: Check what $T$ maps the boundary of $W$ to. From there conclude what $W$ is mapped to. – Martin R Sep 16 '16 at 12:37
• $T$ maps $(\infty,1,-1)$ to $(1,0, \infty)$, the direction of the real axis has changed, By orientation principle, wouldn't it map $W$ upside down? – Leonardo Sep 18 '18 at 0:25
• @Leonardo: I am not sure what you mean. The line segment from $\infty$ to $1$ on the positive real axis is mapped to the segment from $1$ to $0$, the arc from $1$ to $-1$ is mapped to the segment from $0$ to $\infty$ on the positive imaginary axis, etc. The domain lies always to the right, when traversing the boundary in this direction. – Martin R Sep 19 '18 at 7:01