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{\bf Question}

Determine all the circles in $\overline{\bf C}$ that are taken to
themselves by the M\"obius transformation $m(z) =
\frac{3z-2}{2z-1}$.  (That is, determine all the circles $A$ in
$\overline{\bf C}$ satisfying $m(A) =A$.)
\medskip

{\bf Answer}

First, find the fixed points and the type of $m$:

$m(z)=\ds\frac{3z-2}{2z-1}$ det$(m)=-3+4=1$, so $m$ is already
normalized.

\un{$\tau(m)=(3-1)^2=4$ and so $m$ is parabolic.}

\bigskip

\un{Fixed point}:

$\begin{array} {l}
m(z)=z\\2z^2-z=3z-2\\2z^2-4z+2=0\\z^2-2z+1=0\\(z-1)^2=0\ \rm{so}\
m(1)=1 \end{array}$

\bigskip

Since the coefficients of $m$ are real, $m(\bf{R})=\bf{R}$. If $A$
is a circle in $\bar{\bf{C}}$ intersecting $\bf{R}$ in two points
(1=fixed point of $m$ and $z_0$), then $m(A) \ne A$ since $m(z_0)
\ne z_0$. If $A$ is a circle in $\bar{\bf{C}}$ which is tangent to
$\bf{R}$ at 1, then $m(A)=A$. So, the circles taken to themselves
by $m$ are $\bar{\bf{R}}$ and any circle in $\bf{C}$ tangent to
$\bf{R}$ at 1.

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