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{\bf Question}

Determine whether the M\"{o}bius transformation
$$m(z)=\frac{4z-5}{2z+3}$$ is parabolic,elliptic or loxodromic,
and determine it's fixed points.If it is elliptic or loxodromic,
determine its multiplier.
\medskip

{\bf Answer}

Use classification by trace squared: first normalize so that
determinant is 1.

det$(m)=12+10==22$, so $m$ normalized is

$$m(z)=\ds\frac{\frac{4}{\sqrt{22}}z-\frac{5}{\sqrt{22}}}
{\frac{2}{\sqrt{22}}z+\frac{3}{\sqrt{22}}}$$

Trace$^2(m)=\left(\ds\frac{4}{\sqrt{22}}+\ds\frac{3}{\sqrt
22}\right)^2=\ds\frac{49}{22}$

since $0 \leq$ trace $^2(m)<4$, $m$ is elliptic.

Find the multiplier from the trace squared.

$(\lambda+\lambda^{-1})^2=$trace$^2(m)=\ds\frac{49}{22}$, where
the multiplier of $m$ is $\lambda^2$.

Then, $\lambda^2+\lambda^{-1}+2=\ds\frac{49}{22}$, so
$\lambda^4+1-\ds\frac{5}{22}\lambda^2=0$

So,
$\lambda^2=\ds\frac{[\frac{5}{22}+\sqrt{(\frac{5}{22})^2-4}]}{2}$
(and $|\lambda^2|=1$).

$\ds\frac{5}{44} \pm \ds\frac{\sqrt{1911}i}{44}$

\bigskip
The fixed points of $m$ are the solutions to $m(z)=z$, namely

$4z-5=z(2z+3)$, on $2z^2-z+5=0$, so

$z=\ds\frac{1}{4}[1 \pm \sqrt{1-40}]=\un{\ds\frac{1}{4}(1 \pm
\sqrt{39}i)}$.
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