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\begin{document}

{\bf Question}

Show that $|\sin z|^2=\sin^2x+\sinh^2y$ where $z=x+iy$ and hence
that, for any positive integer $n$,

\begin{itemize}
\item[i)]
on the lines $y=\pm(n+\frac{1}{2}), \hspace{0.3in} |\csc\pi
z|^2\leq{\rm csch}^2(\frac{\pi}{2})$

\item[ii)]
on the lines $x=\pm(n+\frac{1}{2}), \hspace{0.3in} |\csc\pi
z|^2\leq1$.

\end{itemize}

The above lines form the sides of a square $\Gamma_n$.  Prove that

$$\lim_{n\to\infty}\int_{\Gamma_n}\frac{\pi\csc\pi z}{z^2}dz=0$$

and, using the calculus of residues, deduce that

$$\sum_{r=1}^\infty\frac{(-1)^{r+1}}{r^2}=\frac{\pi^2}{12}.$$



\vspace{0.25in}

{\bf Answer}

$\sin(x+iy)=\sin x\cos iy+\cos x+\sin iy=\sin x\cosh y+i\cos
x\sinh y$

$|\sin z|^2=\sin^2x\cosh^2y+\cos^2x\sinh^2y$

$\hspace{0.5in}=\sin^2x\cosh^2y+(1-\sin^2x)\sinh^y$

$\hspace{0.5in}=\sin^2x(\cosh^2y-\sinh^2y)+\sinh^2y=\sin^2x+\sinh^2y$

Now if $y=\pm(n+\frac{1}{2}), \hspace{0.2in} |\sin\pi
z|^2\geq\sinh^2\pi y$

$=\sinh^2(\pm(n+\frac{1}{2})\pi\geq\sinh^2\frac{1}{2}\pi$.  So
$|\csc\pi z|^2\leq{\rm csch}^2(\frac{1}{2}\pi)$

If $x=\pm(n+\frac{1}{2}), \hspace{0.2in} |\sin^2\pi
z|\geq\sin^2\pi x=\sin^2(\pm(n+\frac{1}{2})\pi)=1$

So $|\csc^2\pi z|\leq1$

Now on the square $\Gamma_n$, $|z|\geq n+\frac{1}{2}$ and
$|csc^2\pi z|\leq\max(1,{\rm csch}^2\frac{1}{2}\pi)=K^2$

So $\ds\left|\int_{\Gamma_n}\frac{\pi\csc\pi
z}{z^2}dz\right|\leq\frac{\pi
k8(n+\frac{1}{2})}{(n+\frac{1}{2})^2}\to0$ as $n\to\infty$.

Now at $z=n\not=0$, $\ds\frac{\pi\csc\pi z}{z^2}$ has residue
$\ds\frac{(-1)^n}{n^2}$.

At $z=0$, $\ds\frac{\pi\csc\pi z}{z^2}$ has a pole of order 3,
with residue $\ds\frac{\pi^2}{3!}$

So $\ds\int_{\Gamma_n}\frac{\pi\csc\pi z}{z^2}=2\pi
i\left(\frac{\pi^2}{3!}+\sum_{-n, \,
n\not=0}^n\frac{(-1)^r}{r^2}\right)$

thus $\ds\sum_{r=1}^\infty\frac{(-1)^{r+1}}{r^2}=\frac{\pi^2}{12}$

\end{document}