<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Let $m(z)$ be a loxodromic M\"obius transformation that does not
fix $\infty$.  The {\em isometric circle} of $m$ is the circle $\{
z\in {\bf C}\: |\: |m'(z)| =1 \}$, where $m'(z)$ is the derivative
of $m(z)$ in terms of $z$.  Prove the following statement, or give
a counterexample: the isometric circle of a loxodromic M\"obius
transformation (not fixing $\infty$) $m(z)$ is always disjoint
from the isometric circle of its inverse $m^{-1}(z)$.

\medskip
\noindent [Hint: use the standard form for a loxodromic M\"obius
transformation, and try calculating a few specific numerical
examples.]
\medskip

{\bf Answer}

The standard form of a loxodromic fixing $x,\ y$ with multiplier
$k^2$ is

$m(z)=\ds\frac{(xk^{-1}-yk)z+xy(k-k^{-1})}{(k-k^{-1})z+xk+yk^{-1}}$

$\left[\rm{set}\ p(z)=\ds\frac{z-x}{z-y}\ \rm{and}\ \ell(z)=k^2z\
\rm{and\ evaluate}\ p^{-1} \ell p(z) \right]$

det$(m)=(x-y)^2$

$m(z)=\ds\frac{\frac{xk^{-1}-yk}{x-y}z+\frac{xy(k-k^{-1})}{x-y}}
{\frac{k^{-1}-k}{x-y}z+\frac{xk-yk^{-1}}{x-y}}$ with det$(m)=1$.

$m'(z)=\ds\frac{1}{\left(\frac{k^{-1}-k}{x-y}z+\frac{xk-yk^{-1}}{x-y}\right)^2}$

$|m'(z)|=1$:

$$\left|\ds\frac{k^{-1}-k}{x-y}z+\ds\frac{xk-yk^{-1}}{x-y}\right|=1$$

\bigskip

$$\un{\left|z+\ds\frac{xk-yk^{-1}}{x-y} \cdot
\ds\frac{x-y}{k^{-1}-k}\right|=\left|\ds\frac{x-y}{k^{-1}-k}\right|}$$

\bigskip

$$\left|z-\left(\ds\frac{yk^{-1}-xk}{k^{-1}-k}\right)\right|=\ds\frac{|x-y|}{|k^{-1}-k|}$$

This is the isometric circle of $m$.

\newpage
$$m^{-1}(z)=\ds\frac{\left(\ds\frac{xk-yk^{-1}}{x-y}\right)z-\ds\frac{xy(k-k^{-1})}{x-y}}
{\left(\ds\frac{k-k^{-1}}{x-y}\right)z+\ds\frac{xk^{-1}-yk}{x-y}}$$
\bigskip
$$(m^{-1})'(z)=\ds\frac{1}{\left(\left(\ds\frac{k-k^{-1}}{x-y}\right)
z+\ds\frac{xk^{-1}-yk}{x-y}\right)^2}$$
\bigskip
$|(m^{-1})'(z)|=1$:

$$\left|\ds\frac{k-k^{-1}}{x-y}z+\ds\frac{xk^{-1}-yk}{x-y}\right|=1$$

$$\left|z+\ds\frac{xk^{-1}-yk}{k-k^{-1}}\right|=\left|\ds\frac{x-y}{k-k^{-1}}\right|$$

Isometric circle of $m^{-1}(z)$.

Distance between centers:

\begin{eqnarray*}
&amp; &amp;
\left|-\ds\frac{(xk^{-1}-yk)}{k-k^{-1}}-\ds\frac{(yk^{-1}-xk)}{k^{-1}-k}\right|\\
&amp; = &amp; \left|\ds\frac{xk^{-1}-yk-yk^{-1}+xk}{k^{-1}-k}\right|\\ &amp; =
&amp; \left|\ds\frac{(x-y)(k+k^{-1})}{k^{-1}-k}\right| \end{eqnarray*}

Sum of radii:

$$\ds\frac{|x-y|}{|k^{-1}-k|}+\ds\frac{|x-y|}{|k^{-1}-k|}=\ds\frac{2|x-y|}{|k^{-1}-k|}$$

\un{Circles are disjoint} if and only if distance between centers
is greater than sum of radii:

$$\left|\ds\frac{(x-y)(k+k^{-1})}{k^{-1}-k}\right|&gt;\ds\frac{2|x-y|}{|k^{-1}-k|}\
\ \rm{i.e.}\ \un{|k+k^{-1}|&gt;2}.$$

There are complex numbers $k$ so that $|k|&gt;1$ and $|k+k^{-1}|&gt;2$
and so there are loxodromic $m(z)$ for which the isometric circles
of $m$ and $m^{-1}$ are not disjoint. (Note that the conditions on
$m$ for this are independent of the fixed points.)

\end{document}
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