Let $Q := \{x + iy : x > 0, y > 0\}$. Does there exists a conformal mapping $\phi$ from $Q$ to the unit disk such that $\phi(1+i) = 1/2$ and $\phi(1+2i) = -1/2$ ?
Here is what I would do :
- The conformal map from $Q$ to the upper-half plane is $z \mapsto z^2$
- The conformal map from the upper-half plane to the unit disk is $z \mapsto \frac{z-i}{z+i}$
- The image of $1+i$ by the composition of these conformal maps is some $\lambda \in \mathbb{D}$. The conformal map $z \mapsto \frac{z-\lambda}{1-\overline{\lambda}z}$ would map $\lambda$ to 0 and $\frac{z+\frac{1}{2}}{1+\frac{1}{2}z}$ would then map 0 to $\frac{1}{2}$.
- $\phi$ is the composition of all those conformal maps and $\phi(1+i) = 1/2$ as we want.
- Now any conformal map from $Q$ to the unit disk mapping $1+i$ to $\frac{1}{2}$ is a unimodular constant times $\phi$. So all I need to check is wheter $\big|\phi(1+2i)\big|$ is equal $\left|-\frac{1}{2}\right| = \frac{1}{2}$ or not.
The last step is daunting and I'm not even sure that it's true. Is there a shorter/easier path to the answer ?
I think it's best to work from both ends to the middle. Let $z_1 = 1+i$ and $z_2 = 1+2i$. And let $w_1 = \frac{1}{2},\, w_2 = - \frac{1}{2}$. Getting one point of each pair to $0$ via a biholomorphic map is easy. Here, we have $z_1^2 = 2i$, so it would be more convenient to use the map $z \mapsto \dfrac{z-2i}{z+2i}$ to map the upper half-plane to the unit disk. So we'd look at
$$\psi \colon z \mapsto \frac{z^2-2i}{z^2+2i}$$
as a biholomorphic map $Q \to \mathbb{D}$ with $\psi(z_1) = 0$. Then using the automorphism $z \mapsto \dfrac{z - w_1}{1-w_1z}$ of the unit disk to map $w_1$ to $0$, the question is whether
$$\lvert \psi(z_2)\rvert = \biggl\lvert \frac{w_2-w_1}{1-w_1w_2}\biggr\rvert = \frac{4}{5}.$$