\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Write down two explicit M\"obius transformations taking the disc
$D =\{ z\in {\bf C}\: |\: |z -2| < 1\}$ to the disc $E = \{ z\in
{\bf C}\: |\: {\rm Im}(z) <0 \}$.
\medskip

{\bf Answer}

Start by taking $\{|z-2|=1\}$ to $\bar{\bf{R}}$:

take 3 points on $A=\{|z-2|=1\}$, such as $z_1=1, z_2=3, z_3=2+i$
and then take $m \in$ M\"ob$^+$ satisfying $m(z_1)=0,
m(z_2)=\infty, m(z_3)=1$:

$$m(z)=\ds\frac{z-1}{z-3} \cdot \ds\frac{2+i-3}{2+i-1} =
\ds\frac{z-1}{z-3} \cdot \ds\frac{-1+i}{1+i}$$

Since $m(A)=\bar{\bf{R}}$, either $m(D)=\bf{H}$ or
$m(D)=E=\{\rm{Re}(z)<0\}$. Test by checking $m(2)$ since $2 \in
D$:

\begin{eqnarray*}
m(z) & = & \ds\frac{2-1}{2-3} \cdot \ds\frac{-1+i}{1+i}\\ & = &
\ds\frac{1-i}{1+i} \cdot \ds\frac{1-i}{1-i}\\ & = &
\ds\frac{-2i}{2} = -i \in E \end{eqnarray*} and so $m(D)=E$ as
desired.

A second one is \un{$p(z)=m(z)+1$}.

\end{document}