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{\bf Question}

Consider the element of arc-length $\frac{1}{2 +5|z|} |{\rm dz}|$
on ${\bf C}$.  Calculate the circumference, with respect to this
element of arc-length, of the Euclidean circle with center $0$ and
radius $r >0$.

\medskip
\noindent Further, write down the integral for the length, with
respect to this element of arc-length, of the square with vertices
at $\pm r \pm ri$ for $r >0$.  Evaluate it if you can.

\medskip

{\bf Answer}

$C_r$ the circle with center 0 and radius $r$. Parametrize $C_r$
by $f(t)=re^{it}\ \ 0 \leq t \leq t\pi$.

Then, \begin{eqnarray*} f'(t) & = & \ds\int_f \ds\frac
{|\,dz|}{2+5|z|}\\ & = & \ds\int_0^{2\pi}
\ds\frac{|f'(t)|\,dt}{2+5|f(t)|}\\ & = & \ds\int_0^{2\pi}
\ds\frac{r \,dt}{2+5r} = \ds\frac{2\pi r}{2+5r}
\end{eqnarray*}

Parametrize the side of the square $S_r$ from $r-ri$ to $r+ri$ by
$f(t)=r+ti,\ -r \leq t \leq r$.

Then, $f'(t)=i$ and so \begin{eqnarray*} \rm{length}(S_r) & = & 4\
\rm{length(side)}\\ & = & 4 \ds\int_f \ds\frac{|\,dz|}{2+5|z|}\\ &
= & 4\ds\int_{-r}^r \ds\frac{|f'(t)| \,dt}{2+5|f(t)|}\\ & = &
4\ds\int_{-r}^r \ds\frac{\,dt}{2+5\sqrt{r^2+t^2}} \end{eqnarray*}

(The fact that $\ds\frac{1}{2+5|z|}|\,dz|$ is invariant by
notations of $\bf{C}$ fixing 0 is what allows us to say that the
length of the square is 4 times the length of one side.)

(I don't know how to evaluate this integral and I haven't yet
found it in a table.)
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