Find a biholomorphic map I'm currently stuck on how to find a biholomorphic map between two subset $A,B\subset \mathbb{C}$. Given by:
$$ A:= \left\{z\in\mathbb{C}\ \big| \ |z|<1 \right\} \\
   B:= \left\{z\in\mathbb{C}\ \middle| \ \Im(z) \ge (\Re(z))^2\right\} $$
It seems that the circles in $A$ must be mapped onto the parabolas.
But I don't see how to do it!
 A: Use the map $w=z^2$ to map the vertical line $\text{Re}(z)=1$ to the parabola with vertex at $z=1$ whose axis of symmetry is the $x$-axis. The map can then be composed with a usual (birational) map between a circle and a line.
A: The boundary circe of $A$ has equation $x^2+y^2=1$. If you use the transformation $x=\frac{2X}{Y+1}$ and $y=\frac{Y-1}{Y+1}$ then the equation of the circle will be transformed into the equation of the parabola $Y=X^2$. Now check that this is given by a holomorphic map.
A: Since I recently came across this problem and got the answer here, here are the details of the answer of @user72694. 
One map that maps the line $Im(z) = 0$ to $Re(z) = 1$ is $\varphi : z\mapsto iz+1$. Since this also maps the standard parabola to the parabola given by @user72694, this yields a biholomorphism $z\mapsto \varphi^{-1}(\varphi(z)^2) = 2z +iz^2$ between the upper half plane and the "parabola domain". 
I choose the biholomorphism between the upper half plane and the parabola domain to be $\psi(z) = \frac{z-i}{z+i}$ (with inverse $z\mapsto \frac{iz-i}{-z+1}$). Then the biholomorphism $A\to B$ comes out as $z\mapsto \frac{z^2-2iz -i}{iz^2 + 2z -1}$.
