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

{\bf Question}

Let $\ell_1$ be the hyperbolic line contained in the Euclidean
line $\{z\in {\bf H}\: |\: {\rm Re}(z) = 2\}$, and let $\ell_2$ be
the hyperbolic line contained in the Euclidean circle with center
$-3$ and radius $8$.  Determine all the elements of ${\rm
M\ddot{o}b}({\bf H})$ taking $\ell_1$ to $\ell_2$.
\medskip

{\bf Answer}

$\ell_1$ has endpoints at infinity $x_1=2, y_1=\infty$

$\ell_2$ has endpoints at infinity $x_2=-11, y_2=5$

\bigskip

Consider $p(z) \in $M\"ob$^+(\bf{H})$.
$p(z)=\ds\frac{11(z-2)+5}{-(z-2)+1}=\ds\frac{11z-17}{-z+3}$

\bigskip

Then, $p(2)=5, p(\infty)=-11$, and det$(p)=16>0$, and so $p(z) \in
$M\"ob$^+(\bf{H})$ and $p(\ell_1)=\ell_2$.

\bigskip

If $m \in $M\"ob$(\bf{H})$, $m(\ell_1)=\ell_2$, then $p^{-1} \circ
m(\ell_1)=\ell_1$, and so $m=p \circ n, n(\ell_1)=\ell_1$.

\bigskip

stab$_{\rm{M\ddot{o}b}(\bf{H})}(\ell_1)$ is generated by:

\bigskip

$$\begin{array}{l} \mu(z)=-\bar z+4\\
\nu(z)=2+\ds\frac{1}{(z-2)}\\ \lambda(z)=az+2(1-a)\ \ (a>0)
\end{array}$$
(\un{no} parabolics)

\bigskip

(conjugate the generators of stab$_{\rm{M\ddot{o}b}(\bf{H})}(I)$
by $z \mapsto z+2$, where $I$ is the positive imaginary axis).

\bigskip

So, the set of all M\"ob$(\bf{H})$ taking $\ell_1$ to $\ell_2$ is
generated by

$$p \circ
\mu(z)=\ds\frac{11(-\bar{z}+4)-17}{\bar{z}-4+3}=\ds\frac{-11\bar{z}+27}{\bar{z}-1}$$

\bigskip

$$p \circ
\nu(z)=\ds\frac{11\nu(z)-17}{-\nu(z)+3}=\ds\frac{-6z+23}{2z-5}$$

\bigskip

$$p \circ \lambda(z)=\ds\frac{11az+22(1-a)-17}{-az-2(1-a)+3}\ \
(a>0)$$
\end{document}