<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}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Let $p = 1 + i$ and $q = 5 + i$ be two points in ${\bf H}$.  Let
$\ell$ be the hyperbolic line segment joining $p$ to $q$.
Determine the hyperbolic length of $\ell$.
\medskip

{\bf Answer}

\begin{center}
$\begin{array}{c}
\epsfig{file=407-6-1.eps, width=50mm}
\end{array}
\begin{array}{c}
(\textrm{note that } \varphi=\pi-\theta\\
\textrm{since }Im(p)=Im(\underline{q}))
\end{array} $
\end{center}

The hyperbolic line segment $\ell$ joining $p=1+i$ to $q=5+i$ is
contained in the euclidean circle with center 3 and radius 2, and
so can be parametrized by

$f:[\theta,\pi-\theta] \longrightarrow {\bf{H}},\
f(t)=3+\sqrt{5}e^{it}\ (f(\theta)=q,f(\pi-\theta)=p).$

$\rm{Im}(f(t))=\sqrt{5} \sin(t)$

$|f'(t)|=|\sqrt{5}ie^{it}|=\sqrt{5}$

\newpage
So, \begin{eqnarray*} \rm{length}_{\bf{H}}(f) &amp; = &amp;
\ds\int_{\theta}^{\pi-\theta} \ds\frac{1}{\sin(t)} \,dt\\ &amp; = &amp;
\left.\ln|\csc(t)-\cot(t)|\right|_{\theta}^{\pi-\theta}\\ &amp; = &amp;
\ln\left|\ds\frac{\csc(\pi-\theta)-\cot(\pi-\theta)}{\csc(\theta)-\cot(\theta)}
\right|\\ &amp; = &amp;
\ln\left|\ds\frac{\csc(\theta)+\cot(\theta)}{\csc(\theta)-\cot(\theta)}
\right| \end{eqnarray*}

$\begin{array}{ll} \sin(\theta)=\ds\frac{1}{\sqrt 5} &amp;
\csc(\theta)=\sqrt 5\\ \cos(\theta)=\ds\frac{2}{\sqrt 5} &amp;
\cot(\theta)=2 \end{array}$

\un{So, length$_{\bf{H}}(f)=\ln \left|\ds\frac{\sqrt 5 +2}{\sqrt 5
- 2}\right|$}.
\end{document}
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