<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}
\parindent=0pt
\begin{document}

{\bf Question}

Show that the following formula holds in ${\bf H}$:
\[ \cosh({\rm d}_{\bf H}(z,w)) = 1 + \frac{|z- w|^2}{2{\rm Im}(z)
{\rm Im}(w)}. \]
\medskip

{\bf Answer}

Start by showing that the RHSide of the equation is invariant
under M\"{o}b$^+({\bf{H}})$: that is, if $m \in$
M\"{o}b$^+({\bf{H}})$, then

$$\ds\frac{|z-w|^2}{2{\rm{Im}}(z){\rm{Im}}(w)}=
\ds\frac{|m(z)-m(w)|^2}{2{\rm{Im}}(m(z)){\rm{Im}}(m(w))}:
(\star)$$

Write $m(z)=\ds\frac{az+b}{cz+d}\ a,b,c,d \in {\bf{R}},\ ad-bc=1$

\un{Then}:
\begin{description}
\item[(1)]
$$\begin{array} {rcl} |m(z)-m(w)|^2 &amp; = &amp; |m'(z)||m'(w)||z-w|^2\\
{} &amp; {} &amp; {}\\ {} &amp; = &amp; \ds\frac{|z-w|^2}{|cz+d|^2|cw+d|^2}\\ {} &amp;
{} &amp; {}\\ {} &amp; {} &amp; {}\\m(z)-m(w) &amp; = &amp;
\ds\frac{az+b}{cz+d}-\ds\frac{aw+b}{cw+d}\\ {} &amp; {} &amp; {}\\ {} &amp; =
&amp; \ds\frac{(az+b)(cw+d)-(aw+b)(cz+d)}{(cz+d)(cw+d)}\\ {} &amp; {} &amp;
{}\\ {} &amp; = &amp; \ds\frac{z-w}{(cz+d)(cw+d)}
\end{array}$$

$m'(z)=\ds\frac{1}{(cz+d)^2}$

\item[(2)]

\begin{eqnarray*}
{\rm{Im}}(m(z)) &amp; = &amp; {\rm{Im}}\left(\ds\frac{az+b}{cz+d} \cdot
\ds\frac{c\bar{z}+d}{c\bar{z}+d}\right)\\ {} &amp; {} &amp; {}\\ &amp; = &amp;
{\rm{Im}}\left(\ds\frac{acz\bar{z}+bc\bar{z}+adz+bd}{(cz+d)(c\bar{z}+d)}\right)\\
{} &amp; {} &amp; {}\\ &amp; = &amp; \ds\frac{{\rm{Im}}(z)}{(cz+d)(c\bar{z}+d)}
\end{eqnarray*}

and we see that $(\star)$ is satisfied.

Now, given $z,w \in {\bf{H}}$, choose $m \in$
M\"{o}b$^+({\bf{H}})$ so that

$(m(z)=i$ and $m(w)=\lambda i (\lambda&gt;1)$, where
$\ln(\lambda)=d_{\bf{H}}(z,w)$: then

\begin{itemize}
\item
$\cosh(d_{bf{H}}(z,w))=\cosh(\ln(\lambda))=
\ds\frac{1}{2}\left(\lambda+\ds\frac{1}{\lambda}\right)=\ds\frac{\lambda^2+1}{2\lambda}$

\item
\begin{eqnarray*} 1+\ds\frac{|z-w|^2}{2{\rm{Im}}(z){\rm{Im}}(w)} &amp;
= &amp; 1 +\ds\frac{|i-\lambda i|^2}{2 \cdot 1 \cdot \lambda}\\ &amp; = &amp;
1 + \ds\frac{(\lambda-1)^2}{2\lambda}\\ &amp; = &amp;
\ds\frac{2\lambda+\lambda^2-2\lambda+1}{2\lambda}\\ &amp; = &amp;
\ds\frac{\lambda^2+1}{2\lambda}\ {\rm{as\ desired}}
\end{eqnarray*}
\end{itemize}
\end{description}
\end{document}
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