Biholomorphic transformation from half disk onto full one? Consider the unit disk $D = \{ z \in \mathbb C :  |z| < 1 \}$ and the upper half disk $S = \{ z \in D : Im(z) > 0 \}$. 

Then our assignment is to find a biholomorphic map $S \to D$.

Of course $z\to z^2$ won't quite work as $[0,\infty)$ isn't contained in the image. Could you give some hints please? A mapping from $D$ to the half-plane $\mathbb H$ would suffice, as I know the Möbius transformations for $\mathbb H \to D$. A plain Möbius transformation wouldn't work either though, as its inverse would take $D$ to a full disk/plane instead of the semidisk.
Thanks
 A: In this proof I'll show you how to find a conformal map from S to the upper half plane. Then you can take the inverse and compose it with a Möbius transform.
Consider the function $g(z)=\dfrac{1}{z+1}$, which is a Möbius transform, therefore it maps circles/lines to circles/lines. Since it has a pole in -1, we conclude that it maps the interval $[-1,1]$ , as well as the upper circle ( it is the set $\partial S\setminus (-1,1)$) to two lines, which intersect orthogonally at the point $g(1)=\frac{1}{2}$. 
By taking the image of different points of S you can see that the image of S via g is the 4th quadrant, translated by 1/2 to the right, i.e. $\{(x,y)\in\mathbb R^2:x>\frac{1}{2},y<0\}$.
As a result, $g(z)-\frac{1}{2}$ maps S to the 4th quadrant.
Continuing, we rotate by $\pi/2$, so $e^{i\frac{\pi}{2}}(g(z)-\frac{1}{2})=i(g(z)-\frac{1}{2})$ maps S to the 1st quadrant.
Finally, by squaring, we can map the 1st quadrant to the upper half plane, in other words:
$$(i(g(z)-\frac{1}{2}))^2=\dfrac{-(z-1)^2}{4(z+1)^2}$$
maps S to the upper half plane.
Note: now compose with $\dfrac{z-i}{z+i}$ which maps the upper half plane to the unit disk , to get a conformal map from S to D. What you need is the inverse of this map.
