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{\bf Question}

Let $E$ be the ellipse in ${\bf C}$ given by the equation
\[ E =\left\{ z\in {\bf C}\: |\:  \frac{3}{4} \left( {\rm Re}(z)
\right)^2 + \frac{5}{4} \left( {\rm Im}(z) \right)^2 = 1 \right\}.
\]

\medskip
\noindent Determine at least three non-trivial elements $m$ of the
general M\"obius group ${\rm M\ddot{o}b}$ satisfying $m(E) =E$.

\medskip

{\bf Answer}

First, note that $E$ is symmetric with respect to both the
$x$-axis and the $y$-axis ($\bf{R}$-axis and imaginary axis) and
so two elements of M\"{o}b taking $E$ to $E$ are $C(z)=\bar{z}$
(reflection in $\bf{R}$) and $B(z)=-\bar{z}$ (reflection in the
imaginary axis). The comparison of $B$ and $C$ is rotation by
$\pi$ fixing $0,\infty$ (i.e. $m(z)=-\bar{z}$), which also takes
$E$ to $E$.  So, there is a $\bf{Z}_2 \oplus \bf{Z}_2$ subgroup of
M\"{o}b generated by $B,\ C$ contained in $G_E=\{m \in \rm{M
ddot{o}b}|m(E)=E\}$.

\bigskip

[\un{no loxodromic takes $E$ to $E$}: if $m$ is loxodromic and
$m(E)=E$, then the fixed points of $m$ are on $E$. This is
probably most easily seen by conjugating so that the fixed points
of $m$ are 0 and $\infty$ and then noting that a curve invariant
under such a loxodromic is either a line through 0 or a curve that
spirals into 0, and the ellipse gets taken to a curve that does
neither.

\bigskip

\un{no parabolic takes $E$ to $E$}; again, if $m$ is parabolic and
$m(E)=E$, then the fixed point of $m$ is on $E$. Conjugating so
that $m(z)=z+1$, note that the curves invariant under $m$ are
precisely the horizontal lines and other periodic curves, and the
ellipse is neither.

\bigskip

\un{no infinite order elliptic takes $E$ to $E$}: if it did, then
$E$ would contain a dense subset of a circle, which would then
force $E$ to be a circle, which it isn't.]
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