# Finding a conformal map from semi disk to upper half plane.

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 $f(z)=(\frac{i}{2}\frac{z-1}{z+1})^2$

• See example 10.10 here – user1337 Oct 28 '13 at 14:40
• Wow thank you for suggesting the link. There are good examples to understand such type of question. @user1337 – Nrsnr Oct 28 '13 at 14:46

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.