Conformal map from complex plane with a cut-out to unit disc with two given values

Question

Find a conformal map from $D = \mathbb{C} \setminus \{z \in \mathbb{R} : z \leq 0\}$ to the unit circle so that $f(0)=-1$ and $f(i)=0$. My first idea was to maybe to apply $f_1(z) = z^2$ to the set $D$ and it would map $D$ to $\mathbb{C} \setminus \{0\}$. Then I could maybe somehow map that set to the unit circle. Another question is that how do I make sure that the two initial conditions are satisfied? Maybe I could find a third condition based on these two and then find a function such as $f(z) = \frac{az+b}{cz+d}$ which would work?

Answer 1

Note that $z^2$ isn't injective, so it's not gonna do what you want.

On the other hand, you have a well-defined square-root on $D$ by extending the usual square-root on $\mathbb{R}$. Denote this map $\sqrt{\cdot}$ which then maps $D$ conformally onto $(0,\infty)\times i\mathbb{R}$, mapping $i$ to $\frac{1+i}{\sqrt{2}}$ and fixing $0$ (when extending by continuity).

Now, from here, we can map to the unit disk by applying the maps $z\mapsto iz$ (mapping onto the upper half-plane) and $z\mapsto \frac{z-i}{z+i}$ (the Cayley Transform, which maps the upper half-plane conformally to the disk and maps $0$ to $-1$). All in all, the composition is given by $z\mapsto \frac{i\sqrt{z}-i}{i\sqrt{z}+i}$. I

Thus, we've found a conformal equivalence between $D$ and the disk $\mathbb{D}$. As it turns out, it already maps $0$ to $-1$, but there is a trick you can apply that works for a generic situation. Note that if $z_0\in \mathbb{D}$ and $e^{i\theta},e^{i\phi}\in S^1$, then

$$ z\mapsto e^{i\psi}\frac{1-\overline{z_0}e^{i\theta}}{e^{i\theta}-z_0}\frac{z-z_0}{1-\overline{z_0}z} $$ is a conformal automorphism of $\mathbb{D}$ which maps $z_0$ to $0$ and $e^{i\theta}$ to $e^{i\psi}$.