<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 $C_s$ denote the hyperbolic circle in the Poincar\'e disc
${\bf D}$ with hyperbolic radius $s$ and hyperbolic center $0$.
Calculate the circumference of $C_s$ as a function of $s$.

\medskip

{\bf Answer}

If the hyperbolic radius is $s$, then the Euclidean radius is
$r=\tanh(\frac{s}{2})$. Parametrize the circle by $f(t)=re^{-it}$
with $0 \leq t \leq 2\pi$.

\begin{eqnarray*} {\rm{length}}_{\bf{D}}(f) &amp; = &amp; \ds\int_0^{2\pi}
\ds\frac{2r \,dt}{1-r^2}\\ &amp; = &amp; \ds\frac{4\pi r}{1-r^2}\\ &amp; = &amp;
\ds\frac{4\pi \tanh(\frac{s}{2})}{1-\tanh^2(\frac{s}{2})} \cdot
\ds\frac{\cosh^2(\frac{s}{2})}{\cosh^2(\frac{s}{2})}\\ &amp; = &amp;
\ds\frac{4\pi
\sinh(\frac{s}{2})\cosh(\frac{s}{2})}{\cosh^2(\frac{s}{2})-\sinh^2(\frac{s}{2})}\\
&amp; = &amp; 4\pi\sinh(\frac{s}{2})\cosh(\frac{s}{2})\\ &amp; = &amp; 2\pi
\sinh(s) \end{eqnarray*}

(Here we use two identities:

\begin{itemize}
\item
$\cosh^2(x)-\sinh^2(x)=1$
\item
$2\sinh(\frac{x}{2})\cosh(\frac{x}{2})=\sinh(x)$)
\end{itemize}
\end{document}
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