\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

Let $f$ have a pole of order 2 at $z=z_0$ and write
$\phi(z)=(z-z_0)^2f(z)$.  Show that the residue of $f$ at $z_0$ is
equal to $\phi'(z_0)$.

Hence, or otherwise, evaluate by contour integration

\begin{itemize}
\item[i)]
$\ds\int_0^\infty\frac{x^2dx}{(1+x^2)^2}$

\item[ii)]
$\ds\int_0^{2\pi}\frac{d\theta}{(2+\cos\theta)^2}.$
\end{itemize}



\vspace{0.25in}

{\bf Answer}

$f(z)$ has a pole of order 2 at $z_0$

So $\ds f(z)=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+g(z)$

where $g(z)$ is analytic near $z_0$.

$(z-z_0)^2f(z)=b_2+b_1(z-z_0)+(z-z_0)^2g(z)$

$\ds\frac{d}{dz}(z-z_0)^2f(z)=b_1+2(z-z_0)g(z)+(z-z_0)^2g'(z)=b_1$
when $z=z_0$.

\begin{itemize}
\item[i)]
Let $\ds f(z)=\frac{z^2}{(1+z^2)^2}$.

Let $\Gamma$ be:


DIAGRAM


$f(z)$ has a pole of order 2 at $z=i$ inside $\Gamma$, with
residue

$\ds\left.\frac{d}{dz}(z-i)^2f(z)\right|_{z=i}$

$\ds=\frac{d}{dz}\frac{z^2}{(z+i)^2}=
\left.\frac{(z+i)^22z-z^22(z+i)}{(z+i)^4}\right|_{z=i}=
-\frac{i}{4}$

So $\ds\int_{\Gamma}f(z)dz=-2\pi i\frac{i}{4}=\frac{\pi}{2}$

On the semi-circle $\ds|f(z)|\leq\frac{|z|^2}{(|z|^2-1)^2}=
\frac{R^2}{(R^2-1)^2}$

So $\ds\left|\int_Cf(z)dz\right|\leq\frac{R^2}{(R^2-1)^2}2\pi
R\to0$ as $R\to\infty$

Thus $\ds\int_{-\infty}^\infty f(x)dx=\frac{\pi}{2}$ and so
$\ds\int_0^\infty\frac{x^2}{(1+x^2)^2}dx=\frac{\pi}{4}$.


\item[ii)]
Let $z=e^{i\theta}$ and $C$ be the unit circle

$\ds\cos\theta=\frac{1}{2}\left(z+\frac{1}{z}\right)$ and $\ds
d\theta=\frac{dz}{iz}$

$\ds I=\int_0^{2\pi}\frac{d\theta}{(2+\cos\theta)^2}=
\int_C\frac{dz}
{iz\left(2+\frac{1}{2}\left(z+\frac{1}{z}\right)\right)^2}=
\frac{4}{i}\int_C\frac{z}{(z^2+4z+1)^2}dz$

Now $z^2+4z+1=0 \begin{array}{l}\ds {\rm if\ }
z=z_0=\frac{-4+2\sqrt3}{2} {\rm \ inside\ }C\\ \ds {\rm or\ }
z=z_1=\frac{-4+-\sqrt3}{2} {\rm \ outside\ }C\end{array}$

So $f(z)$ has a pole of order 2 at $z=z_0$, with residue

$\ds\left.\frac{d}{dz}\frac{z}{(z-z_1)^2}\right|_{z=z_0}=
\frac{(z_0-z_1)^2-2z_0(z_0-z_1)}{(z_0-z_1)^4}$

$\ds=\frac{(2\sqrt3)^2-(-4+2\sqrt3)2\sqrt3}{(2\sqrt3)^4}=
\frac{\sqrt3}{18}$

So $\ds I=\frac{4}{i}2\pi i\frac{\sqrt3}{18}=
\frac{4\sqrt3}{9}\pi$

\end{itemize}

\end{document}