D = {$z \in \mathbb{C}: |z| < 1 $} and $\Omega$ = {$z \in \mathbb{C}: |z-\sqrt3i| < 2$} $\cap $ {$z \in \mathbb{C}: |z+\sqrt3i| < 2$}. I want to get the explicit solution of this conformal mapping from D to $\Omega$.
This is how I started:
To solve it I used the bilinear transformation $w = \frac{az+b}{cz+d}$.
The points of z I selected were 1, -1, and i (since it is the unit circle). Then from $\Omega$, I must select other three points.
Here is my question: is it okay to choose 1, -1, i from $\Omega$ too? Thanks.
Such a bilinear (or Möbius) transformation will always map circles (such as $\partial D$) to circles (or lines), but the boundary $\partial \Omega$ of $\Omega$ is lens-shaped, with two $60^\circ$ vertices at $-1$ and $1$. So you do need something more.
I would start at $\Omega$, use a bilinear map to transport one of the vertices (e.g., $-1$) to infinity ($z\mapsto \frac1{z+1}$); this turns the two circular arcs of $\partial \Omega$ into lines intersecting at still the same angle of $60^\circ$ at the image point of the other vertex $1$, i.e., at $\frac12$. Now transport this to the origin ($z\mapsto z-\frac12$) and raise to a suitable power to turn the 60° into "smooth" 180° ($z\mapsto {z^3}$). In total, this turns $\Omega$ into a half plane, from where you probably know how to arrive at the unit disk.
In total, thus gives you $\Omega \to D$. To obtain $D\to \Omega$, try to invert the function.