Linear Map that maps the disk of radius 2 to the lower half plane

Question

Is their a general procedure to find a linear transformation that maps $\Delta(n)$, the disc of radius $n\in\mathbb{N}$, to the lower half plane? For example, how can I find linear map that maps $\Delta(2)$ to the lower half plane of $\mathbb{C}$? I am aware that I somehow have to use the cross ratio and define the map at three points but still don't understand how one achieves this.

Answer 1

Well, people do use the cross-ratio, but I just do it by guess and by golly. Here’s my method:

I’m going to try to send the topmost point $ni$ of the boundary of the disk to the origin, and the two real points $-n$ and $n$ on the boundary to $-1$ and $1$, respectively. I’m going to hope that $-ni$ goes to $\infty$, too.

Just dealing with $ni\mapsto0$ gives me \begin{equation}z\mapsto\frac{z-ni}{Cz+D}\,,\end{equation}and I want to choose $C$ and $D$ so that the two real points go to $\pm1$. \begin{matrix} \text{For }-1:&\frac{-n-ni}{-nC+D}&=-1&\text{so}&nC-D=&-n-ni\\ \text{For }1:&\frac{n-ni}{nC+D}&=1&\text{so}&nC+D=&n-ni \end{matrix} Thus we get $C=-i$, and $D=n$. Our fractional-linear map should be \begin{equation}z\mapsto\frac{z-ni}{-iz+n}\,,\end{equation} in which the denominator does indeed become zero under $z=-in$. Just to check, let’s see where the origin goes: $0\mapsto-ni/n=-i$, which indeed is in the lower half-plane. That’s it!