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{\bf Question}

Consider the homeomorphism $f: {\bf H}\rightarrow {\bf H}$ given
by $f(z) = z  + {\rm Re}(z)$.  Prove that the hyperbolic length of
a path in ${\bf H} $ (measured with respect to the element of
arc-length $\frac{1}{{\rm Im}(z)} \: |{\rm dz}|$) is not invariant
under $f$.

\medskip
\noindent Does there exist any non-zero real valued function $g$
so that the hyperbolic length of a path in ${\bf H} $ (measured
with respect to the element of arc-length $\frac{1}{{\rm Im}(z)}
\: |{\rm dz}|$) is not invariant under $f(z) = z + g({\rm
Re}(z))$?
\medskip

{\bf Answer}

Consider $f+o:[0,1] \longrightarrow {\bf H},\ f_0(t)=t+(1+t)i\ \
f_0'(t)=1+i$

\bigskip

length$_{\bf H}(f_0)=\ds\int_0^1 \ds\frac{\sqrt 2}{1+t} \,dt =
\sqrt 2 \ln(2)$.

\bigskip

$f \circ f_0(t)=f_0(t)+\rm{Re}(f_0(t))=2t+(1+t)i\ \ (f \circ
f_0)'(t)=2+i$.

\bigskip

length$_{\bf H}(f \circ f_0)=\ds\int_0^1 \ds\frac{\sqrt 5}{1+t}
\,dt = \sqrt{5}\ln(t) \ne {\rm{length}}_{\bf H}(f_0)$.

\bigskip

Suppose now that there exists $g:{\bf R} \longrightarrow {\bf R}$
so that, for all paths $f_0:[a,b] \longrightarrow {\bf H}$ so that

$${\rm{length}}_{\bf H}(f_0)={\rm{length}}_{\bf H}(f \circ f_0)$$

\bigskip

Write \begin{eqnarray*} f_0(t) &amp; = &amp; x(t)+iy(t)\\ f \circ f_0(t) &amp;
= &amp; x(t)+g(x(t))+iy(t) \end{eqnarray*}

\bigskip

\hspace{0.5in} length$_{\bf H}(f_0)=\ds\int_a^b \ds\frac{1}{y(t)}
\sqrt{(x'(t))^2+(y'(t))^2} \,dt$

\hspace{0.5in} length$_{\bf H}(f\circ f_0)=\ds\int_a^b
\ds\frac{1}{y(t)} \sqrt{(x'(t))^2(1+s'(x(t)))^2+(y'(t))^2} \,dt$

\bigskip

consider $f_0$ with $y(t)=1$ so that $y'(t)=0$, hence

$$\int_a^b \sqrt{(x'(t))^2} \,dt = \int_a^b
\sqrt{(x'(t))^2(1+g'(x(t)))^2} \,dt$$

\bigskip

using an argument similar to the one given in class, since this is
true for all intervals $[a,b] \subseteq {\bf R}$ and all $g$, we
have that $g'(x)=0$ all $x$ and so $g$ is constant (and $g$
constant is a M\"{o}bius transformation).

\bigskip

[Actually, have $(1+g'(x(t)))^2=1$ so either

\bigskip

$1+g'(x(t))=1$ in which case $g'(x(t))=0$ and so $g$ is constant,
or

\bigskip

$1+g'(x(t))=-1$ so $g'(x(t))=-2$ and so $g(x)=2x+c$

\bigskip

So, either $g$ is constant (M\"{o}bius transformation) or
$g(x)=-2x+c$ in which case $z+g({\rm{Re}}(z))=-\bar z +c$ again in
M\"{o}b(${\bf H}$).]
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