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{\bf Question}

 Let $D_s$ be the hyperbolic disc in the Poincar\'e disc ${\bf
D}$ with hyperbolic radius $s$, and let $C_s$ be the hyperbolic
circle with hyperbolic radius $s$ that bounds $D_s$.  Describe the
behavior of the quotient
\[ q(s) =\frac{{\rm length}_{\bf D}(C_s)}{{\rm area}_{\bf D}(D_s)} \]
as $s\rightarrow 0$ and as $s\rightarrow\infty$.

\medskip
\noindent Compare the behavior of $q$ with the analogous quantity
calculated using a Euclidean disc and a Euclidean circle.
\medskip

{\bf Answer}

We know from exercise sheet 8 that
length$_{\bf{D}}({\bf{C}}_s)=2\pi \sinh(s)$.

To calculate area$_{\bf{D}}({\bf{D}}_s)$:

Recall that the euclidean radius of ${\bf{D}}_s$ is
$R=\tanh(\frac{1}{2}s)$, and so

\begin{eqnarray*} area_{\bf{D}}({\bf{D}}_s) & = & \ds\int_0^{2\pi}
\! \ds\int_0^R \ds\frac{4}{(1-|z|^2)^2} \,dx\,dy\\ & = &
\ds\int_0^{2\pi} \! \ds\int_0^R \ds\frac{4r \,dr
\,d\theta}{(1-r^2)^2}\\ & = & 8\pi\ds\int_0^R \ds\frac{r
\,dr}{(1-r^2)^2}\\ & = & 8\pi
\left.\ds\frac{1}{2}(1-r^2)\right|_0^R\\ & = &
8\pi\left(\ds\frac{1}{2(1-R^2)}-\ds\frac{1}{2}\right)\\ & = &
\ds\frac{4\pi}{1-R^2}-4\pi\\ & = &
\ds\frac{4\pi(1-1+R^2)}{1-R^2}\\ & = &
\ds\frac{4\pi\tanh^2(\frac{1}{2}s)}{1-\tanh^2(\frac{1}{2}s)}\\ & =
& 4\pi\sinh^2(\frac{1}{2}s) \end{eqnarray*}

\newpage
and so \begin{eqnarray*}
q(s)=\ds\frac{\rm{length}_{\bf{D}}(C_s)}{\rm{area}_{\bf{D}}(D_s)}
& = & \ds\frac{2\pi\sinh(s)}{4\pi\sinh^2(\frac{1}{2}s)}\\ & = &
\ds\frac{4\pi\sinh(\frac{1}{2}s)\cosh(\frac{1}{2}s)}{4\pi\sinh^2(\frac{1}{2}s)}\\
& = & \ds\frac{\cosh(\frac{1}{2}s)}{\sinh(\frac{1}{2}2)}\\ & = &
\ds\frac{e^{\frac{1}{2}s}+e^{\frac{-1}{2}s}}{e^{\frac{1}{2}s}-e^{\frac{-1}{2}s}}\\
& = & \ds\frac{e^s +1}{e^s -1} \end{eqnarray*}

as $s \to \infty,\ q(s) \to 1$.

as $s \to 0^+,\ q(s) \to \infty$ (since numerator $\to 2$ and
denominator $\to 0$.)

\bigskip

The analagous euclidean quantity is

$$q_E(s)=\ds\frac{\rm{length}_{\bf{D}}(C_s)}{\rm{area}_{\bf{D}}(D_s)}=\ds\frac{2\pi
s}{\pi s^2}=\ds\frac{2}{s}$$

as $s \to \infty,\ q_E(s) \to 0$ (different from $q(s)$).

as $s \to 0^+,\ q_E(s) \to \infty$ (as $q(s)$)

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