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\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Calculate the hyperbolic distance in ${\bf H}$ between $z = 5i$
and $w = e^{i\theta}$, where $0 <\theta <\pi$.  For what value (or
values) of $\theta$ (if any) is this distance minimized?
\medskip

{\bf Answer}

\un{$p=5i,q=e^{i\theta}=\cos(\theta)+i\sin(\theta)$}

The euclidean line segment joining $p$ and $q$ has midpoint
$\ds\frac{1}{2} \cos(\theta)+i\ds\frac{1}{2}(5+\sin(\theta))$ and
slope $m=\ds\frac{\sin(\theta)-5}{\cos(\theta)}$ and so the
perpendicular bisector has equation

$$y-\ds\frac{1}{2}(5+\sin(\theta)0=\ds\frac{\cos(\theta)}
{5-\sin(\theta)}(x-\frac{1}{2}\cos(\theta))$$

\un{Setting $y=0$}:

$$\ds\frac{\sin(\theta)-5}{2\cos(\theta)}(5+\sin(\theta))=
x-\ds\frac{1}{2}\cos(\theta)$$

$$\ds\frac{\sin^2(\theta)-25}{2\cos(\theta)}+\ds\frac{1}{2}\cos(\theta)=x\
\ \ \ {\rm{so}}\ x=\ds\frac{-12}{\cos(\theta)}$$

\bigskip

The center of the euclidean circle containing the line through $p$
and $q$ is $c=\ds\frac{-12}{\cos(\theta)}$ and the radius is
$r=|c-5i|$.

\bigskip

Suppose now that $0 < \theta \leq \frac{\pi}{2}$ (if
$\theta>\frac{\pi}{2}$, then we use that
$d_{\bf{H}}(e^{i\theta},5i)=d_{\bf{H}}(B(e^{i\theta}),
B(5i))=d_{\bf{H}}(e^{i(\pi-\theta)},5i)$ and $0 < \pi-\theta \leq
\frac{\pi}{2}$).

\begin{center}
$ \begin{array}{c}
\epsfig{file=407-8-1.eps, width=60mm}
\end{array}
\begin{array}{c}
sin(\alpha)=\ds\frac{\sin(\theta)}{r}\\
sin(\beta)=\ds\frac{5}{r}
\end{array} $
\end{center}

\bigskip

Parametrize the hyperbolic line segment between $5i$ and
$e^{i\theta}$ by $f(t)=re^{it}+c,\ \alpha \leq t \leq \beta$.
($0<\frac{\pi}{2}$)

\bigskip

\begin{eqnarray*} d_{\bf{H}}(e^{i\theta},5i) & = &
\rm{length}_{\bf{H}}(f)\\ & = & \ds\int_{\alpha}^{\beta}
\ds\frac{1}{\sin(s)} \,ds\\ & = &
\ln\left|\ds\frac{\csc(\beta)-\cot(\beta)}{\csc(\alpha)-\cot(\alpha)}\right|
\end{eqnarray*}

\bigskip

$\begin{array} {ll}
\cos(\alpha)=\ds\frac{\frac{12}{\cos(\theta)+\cos(\theta)}}{r} &
\sin(\alpha)=\ds\frac{\sin(\theta)}{r}\\\cos(\beta)=\ds\frac{12}{\cos(\theta)r}
& \sin(\beta)=\ds\frac{5}{r} \end{array}$

\bigskip

$\csc(\beta)-\cot(\beta)=\ds\frac{r}{5}-\ds\frac{12}{5\cos(\theta)}=
\ds\frac{r\cos(\theta)-12}{5\cos(\theta)}$

\bigskip

$\csc(\alpha)-\cot(\alpha)=\ds\frac{r}{\sin(\theta)}-
\ds\frac{\frac{12}{\cos(\theta)}+\cos(\theta)}{\sin(\theta)} =
\ds\frac{r\cos(\theta)-12-\cos^2(\theta)}{\sin(\theta)\cos(\theta)}$

\bigskip

\begin{eqnarray*}
\ds\frac{\csc(\beta)-\cot(\beta)}{\csc(\alpha)-\cot(\alpha)}& = &
\ds\frac{(r\cos(\theta0-12)\sin(\theta)}{5(r\cos(\theta)-12-\cos^2(\theta))}\\
r & = & \sqrt{\ds\frac{144}{\cos^2(\theta)}+25}\\ r\cos(\theta) &
= & \sqrt{144+25\cos^2(\theta)} \ \ (\rm{since}\ \cos(\theta)>0).
\end{eqnarray*}

\bigskip

$\ds\frac{\csc(\beta)-\cot(\beta)}{\csc(\alpha)-\cot(\alpha)}=\ds\frac{(
\sqrt{144+25\cos^2(\theta)}-12)\sin(\theta)}{5(\sqrt{144+25\cos^2(\theta)}-12-\cos^2(\theta))}$

\end{document}
