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

{\bf Question}

\begin{itemize}
\item[a)]
Let $f(z)$ have a pole of order 2 at $z=z_0$, and write
$g(z)=(z-z_0)^2f(z)$.  Show that the residue of $f$ at $z_0$ is
$g'(z_0)$.

Evaluate, by contour integration

$$\int_{-\infty}^\infty\frac{x^2}{(x^2+a^2)^2}dx$$

where $a$ is a positive real number.


\item[b)]
Evaluate, by contour integration

$$\int_0^{2\pi}\frac{d\theta}{8\cos^2\theta+1}.$$

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$f(z)$ has a pole of order 2 at $z=z_0$, so the Laurent expansion
in a neighbourhood of $z_o$ gives

$\ds f(z)=\frac{\alpha}{(z-z_0)^2}+\frac{\beta}{(z-z_0)}+\phi(z)$

where $g(z)$ is analytic in the neighbourhood.

So $(z-z_0)^2f(z)=\alpha+\beta(z-z_0)+(z-z_0)^2\phi(z)$

$\ds\frac{d}{dz}((z-z_0)^2f(z))=\beta+2(z-z_0)g(z)+(z-z_0)^2\phi(z)$

Putting $z=z_0$ gives the result.

${}$

Let $\ds f(z)=\frac{z^2}{(z^2+a^2)^2}$, this has a pole of order 2
at $z=ia$ in the upper half plane.

Its residue is given by

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

$\ds\left.\frac{(z+ia)^22z-z^22(z+ia)}{(z+ia)^4}\right|_{z=ia}=
\frac{4(ia)^3}{16(ia)^4}=\frac{1}{4ia}$

We integrate $f(z)$ round the contour $\Gamma$ comprising the line
segment $(-R,R)$ on the real axis, and the semicircle
$C=\{z=Re^{it} \,\,\, 0\leq t\leq\pi\}$ where $R>a$.

Then on $\ds C \hspace{0.5in}
|f(z)|=\frac{|z|^2}{|z^2+a^2|^2}\leq\frac{r^2}{(R^2-a^2)^2}$

and so $\ds\left|\int_C
f(z)dz\right|\leq\frac{R^2}{(R^2-a^2)^2}\pi R\to0$ as
$R\to\infty$.

Now $\ds\int_\Gamma f(z)dz=2\pi i\frac{1}{4ia}=\frac{\pi}{2a}$

and $\ds\int_\Gamma f(z)dz=\int_C f(z)dz+\int_{-R}^R f{x}dx$

Letting $R\to\infty$ gives $\ds\int_{-\infty}^\infty
f(x)dx=\frac{\pi}{2a}$


\item[b)]
Let $z=e^{i\theta}$ and $C$ be the unit circle, then
$dz=ie^{i\theta}d\theta$ so $d\theta=\frac{dz}{iz}$

$\ds\cos\theta=\frac{1}{2}\left(z+\frac{1}{z}\right)$.  Hence we
have

$\ds I=\int_0^{2\pi}\frac{d\theta}{8\cos^2\theta+1}= \int_C
\frac{dz}{iz\left(8\frac{1}{4}\left(z+\frac{1}{z}\right)^2+1\right)}$

$\ds=\frac{1}{i}\int_C\frac{dz}{z\left(2z^2+4+\frac{2}{z^2}+1\right)}
=\frac{1}{i}\int_C\frac{zdz}{2z^4+5z^2+2}$

$\ds=\frac{1}{2i}\int_C\frac{zdz}
{\left(z^2+\frac{1}{z}\right)\left(z^2+2\right)}$

Let $\ds f(z)=\frac{z}
{\left(z^2+\frac{1}{z}\right)\left(z^2+2\right)}$

Then $f(z)$ has simple poles at $z=\pm\frac{i}{\sqrt2}$ inside
$C$.

Res$\ds\left(f,\frac{i}{\sqrt2}\right)=
\lim_{z\to\frac{i}{\sqrt2}}\left(z-\frac{i}{\sqrt2}\right)f(z)=
\lim_{z\to\frac{i}{\sqrt2}}\frac{z}{\left(z+\frac{i}{\sqrt2}\right)(z^2+2)}$

$\ds=\frac{\frac{i}{\sqrt2}}{\frac{2i}{\sqrt2}\frac{3}{2}}=\frac{1}{3}$

Res$\ds\left(f,\frac{-i}{\sqrt2}\right)=
\lim_{z\to-\frac{i}{\sqrt2}}\left(z+\frac{i}{\sqrt2}\right)f(z)=
\lim_{z\to-\frac{i}{\sqrt2}}\frac{z}{\left(z-\frac{i}{\sqrt2}\right)(z^2+2)}$

$\ds=\frac{-\frac{i}{\sqrt2}}{-\frac{2i}{\sqrt2}\frac{3}{2}}=\frac{1}{3}$

So $\ds I=\frac{1}{2i}\int_C f(z)dz=\frac{1}{2i}2\pi
i\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{2\pi}{3}$

\end{itemize}

\end{document}
