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

{\bf Question}

For each of the following M\"obius transformations, determine the
fixed points, normalize so that the determinant is $1$, and
determine the type (parabolic, elliptic, loxodromic) of the
transformation.  If the transformation is elliptic or loxodromic,
determine its multiplier.
\begin{itemize}
\item[(a) ] $m(z) = \frac{3z-3}{z+2}$;
\item[(b) ] $m(z) = \frac{-z-16}{2z+2}$;
\item[(c) ] $m(z) = 8z + 3+ i$;
\item[(d) ] $m(z) = \frac{4}{7z+2i}$;
\end{itemize}
\medskip

{\bf Answer}

\begin{description}
\item[(a)]
$m(z)=\ds\frac{3z-3}{z+2}$,\ \ det$(m)=9$, so m \un{normalized} is
$m(z)=\ds\frac{z-1}{\frac{1}{3}z+\frac{2}{3}}$ (divide all
coefficients by $\sqrt{\rm{det}(m)}=3$)

\un{fixed points}:

$z = \ds\frac{z-1}{\frac{1}{3}z+\frac{2}{3}}$

$\Rightarrow \ds\frac{1}{3}z^2+\ds\frac{2}{3}z = z-1$

$\Rightarrow \ds\frac{1}{3}z^2-\ds\frac{1}{3}z+1=0$

$\Rightarrow z^2-z+3=0$

$\Rightarrow z=\ds\frac{1}{2}(1 \pm \sqrt{1-12})$

$\Rightarrow \un{z=\ds\frac{1}{2}(1 \pm \sqrt{11}i)}$

$\tau(m)=\left(1+\ds\frac{2}{3}\right)^2=\ds\frac{25}{9}<4$ and so
\un{$m$ is elliptic}.

\newpage
\un{multiplier is $\lambda^2$}, where
$(\lambda+\lambda^{-1})^2=\tau(m)=\ds\frac{25}{9}$

$$\begin{array} {rcl} \lambda^2+2+\lambda^{-2} & = &
\ds\frac{25}{9}\\ \lambda^2-\ds\frac{7}{9}+\lambda^{-2} & = & 0\\
\lambda^4-\ds\frac{7}{9}\lambda^2+1 & = & 0 \end{array}$$

$$\un{\lambda^2=\ds\frac{1}{2}\left(\ds\frac{7}{9} \pm
\sqrt{\left(-\ds\frac{7}{9}\right)^2-4}\right)}$$

(Can choose either.)

\item[(b)]
$m(z)=\ds\frac{-z-16}{2z+2}$,\ \ det$(m)=30$, so m \un{normalized}
is

$$m(z)=\ds\frac{\frac{-1}{\sqrt{30}}z-\frac{16}{\sqrt{30}}}{\frac{2}
{\sqrt{30}}z+\frac{2}{\sqrt{30}}}$$

\un{fixed points}:

$z = \ds\frac{-z-16}{2z+2}$

$\Rightarrow 2z^2+2z = -z-16$

$\Rightarrow 2z^2+3z+16=0$

$\Rightarrow z=\ds\frac{1}{4}(-3 \pm \sqrt{9-128})$

$\Rightarrow \un{z=\ds\frac{1}{4}(-3 \pm \sqrt{119}i)}$

$\tau(m)=\left(-\ds\frac{1}{\sqrt{30}}+\ds\frac{2}{\sqrt{30}}\right)^2=\ds\frac{1}{30}<4$
and so \un{$m$ is elliptic}.

\un{multiplier is $\lambda^2$}, where
$(\lambda+\lambda^{-1})^2=\tau(m)=\ds\frac{1}{30}$

$$\begin{array} {rcl} \lambda^2+2+\lambda^{-2} & = &
\frac{1}{30}\\ \lambda^2+\frac{59}{30}+\lambda^{-2} & = & 0\\
\lambda^4+\frac{59}{30}\lambda^2+1 & = & 0 \end{array}$$

$$\un{\lambda^2=\ds\frac{1}{2}\left(-\ds\frac{59}{30} \pm
\sqrt{\left(\ds\frac{59}{30}\right)^2-4}\right)}$$

\item[(c)]
$m(z)=8z+3+i$,\ \ det$(m)=8$, so m \un{normalized} is
$m(z)=\ds\frac{\sqrt{8}z+\frac{3+i}{\sqrt{8}}}{0z+\frac{1}{\sqrt{8}}}$

\un{fixed points}: one is \un{$\infty$}. The other is the solution
of $z=8z+3+i \Rightarrow 7z=-3-i \Rightarrow z=\ds\frac{-3-i}{7}$

$\tau(m)=\left(\sqrt{8}+\ds\frac{1}{\sqrt{8}}\right)^2$

$=8+2+\ds\frac{1}{8}=\ds\frac{81}{8}>4$ and so \un{$m$ is
loxodromic}.

\un{multiplier is $\lambda^2$}, where
$(\lambda+\lambda^{-1})^2=\tau(m)=\ds\frac{81}{8}$

$$\begin{array} {rcl} \lambda^2+2+\lambda^{-2} & = &
\ds\frac{81}{8}\\ \lambda^2+\ds\frac{-65}{8}+\lambda^{-2} & = &
0\\ \lambda^4-\ds\frac{65}{8}\lambda^2+1 & = & 0 \end{array}$$

$$\un{\lambda^2=\ds\frac{1}{2}\left(\ds\frac{65}{8} \pm
\sqrt{\left(\ds\frac{-65}{8}\right)^2-4}\right)=8 {\rm \ or \ }
\ds\frac{1}{8}}$$

\item[(d)]
$m(z)=\ds\frac{4}{7z+2i}=\ds\frac{0z+4}{7z+2i}$,\ \ det$(m)=-28$,
so m \un{normalized} is
$$m(z)=\ds\frac{\frac{4i}{\sqrt{28}}}{\frac{7i}{\sqrt{28}}z+\frac{-2}{\sqrt{28}}}$$

\un{fixed points}:

$z = \ds\frac{4}{7z+2i}$

$\Rightarrow 7z^2+2iz = 4$

$\Rightarrow 7z^2+2iz-4=0$

$\Rightarrow z=\ds\frac{1}{14}(-2i \pm \sqrt{-4+112})$

$\Rightarrow \un{z=\ds\frac{1}{14}(-2i \pm \sqrt{108})}$

$\tau(m)=\left(-\ds\frac{2}{\sqrt{28}}\right)^2=\ds\frac{4}{28}=\ds\frac{1}{7}<4$
and so \un{$m$ is elliptic}.

\un{multiplier is $\lambda^2$}, where
$(\lambda+\lambda^{-1})^2=\tau(m)=\ds\frac{1}{7}$

$$\begin{array} {rcl} \lambda^2+2+\lambda^{-2} & = &
\ds\frac{1}{7}\\ \lambda^2+\ds\frac{13}{7}+\lambda^{-2} & = & 0\\
\lambda^4+\ds\frac{13}{7}\lambda^2+1 & = & 0 \end{array}$$

$$\un{\lambda^2=\ds\frac{1}{2}\left(-\ds\frac{13}{7} \pm
\sqrt{\left(\ds\frac{13}{7}\right)^2-4}\right)}$$

\end{description}

\end{document}