Find a conformal mapping $f$ of semi-disk$S=\{z: \vert z\vert \lt 1, Im z\gt 0\}$ onto the upper plane.
Again I used composition of conformal map.
First of all, let's define a conformal map $f_1$ from semi-unit disk to unit disk. But how?
Secondly let's define another conformal map $f_2$ from unit disk to upper half-plane by möbius transformation $$T=\frac{z-1}{z+1}$$
And $T(0)=-1$
By mutiplying with $-i$, I get $f_2=-i\frac{z-1}{z+1}$
And composition of them: $$f(z)=f_2 of_2$$
Then solution hint gives $f(z)=(\frac{i}{2}\frac{z-1}{z+1})^2$
Best Answer
A real time-saver is the Joukowski map $j(z)=\frac12(z+z^{-1})$ which maps the exterior of the unit disk conformally onto the complement of the segment $[-1,1]$. In addition, the signs of real and imaginary part are preserved by $j$. It takes some effort to establish these properties, but once they are available, a number of routine mapping exercises can be done quickly.
In your case, $z\mapsto 1/z$ transforms the domain into the lower half of $\{z:|z|>1\}$. As mentioned above, $j$ maps that domain onto the lower halfplane. Hence, the composition $-j(1/z)$ does what you want.