<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Suppose $f(z)=p(z)/q(z)$ where $p(z_0)\not=0, \,\,\, q(z_0)=0$ and
$q'(z_0)\not=0$.  Show that $f(z)$ has a simple pole at $z_0$ with
residue $p(z_0)/q'(z_0)$.

Evaluate the following integrals.  Justify any limits used.

\begin{itemize}
\item[i)]
$\ds\int_0^\infty\frac{x^2dx}{(x^2+1)(x^2+4)}$

\item[ii)]
$\ds\int_0^{2\pi}\frac{d\theta}{9\cos^2\theta+4\sin^2\theta}$

\end{itemize}



\vspace{0.25in}

{\bf Answer}

$\ds f(z)=\frac{p(z)}{q(z)}$

$\ds(z-z_0)f(z)=p(z)\frac{(z-z_0)}{q(z)-q(z_0)}\rightarrow
\frac{P(z_0}{q'(z_0)}$ as $z\to z_0$

Thus $f(z)$ has a simple pole at $z_0$ with residue
$\ds\frac{P(z_0)}{q'(z_0)}$.

\begin{itemize}
\item[i)]
We integrate $\ds
f(z)\frac{z^2}{(z^2+1)(z^2+4)}=\frac{z^2}{z^4+5z^2+4}=
\frac{p(z)}{q(z)}$

The contour used is $C=\{-R,R\}+\{Re^{it} \,\,\, 0\leq
t\leq\pi\}$. Within this the singularities are at $z=i$ and
$z=2i$.

We find the residues:

$\ds z=i \hspace{0.5in} {\rm res}=
\left.\frac{z^2}{4z^3+10z}\right|_{z=i}=\frac{-1}{-4i+10i}=
\frac{-1}{6i}=\frac{i}{6}$

$\ds z=2i \hspace{0.4in} {\rm res}=
\left.\frac{z^2}{4z^3+10z}\right|_{z=2i}=\frac{-4}{-32i+20i}=
\frac{1}{3i}=\frac{-i}{3}$

So, provided $\ds R&gt;2, \,\,\, \int_C f(z)dz=2\pi
i\left(\frac{-i}{6}\right)=\frac{\pi}{3}$

Now for $\ds|z|=R, \,\,\, |f(z)|=
\frac{R^2}{|z^2+1||z^2+4|}\leq\frac{R^2}{(R^2-1)(R^2-4)}$

So for the semi-circular part $C_1$ of the contour,

$\ds\int_{C_1}f(z)dz\leq\frac{R^2\pi R}{(R^2-1)(R^2-4)}\to0$ as
$R\to\infty$.

Thus, letting $R\to\infty$, we obtain $\ds\int_{-\infty}^\infty
f(x)dx=\frac{\pi}{3}$, so $\ds\int_0^\infty f(x)dx=\frac{\pi}{6}$

\item[ii)]
Let $z=e^{i\theta}$.  So $\cos\theta=\frac{1}{2}(z+\frac{1}{z}),
\,\,\, \sin\theta=\frac{1}{2i}(z-\frac{1}{z}), \,\,\,
d\theta=\frac{dz}{iz}$

So the integral becomes

$\ds\int_C\frac{dz}
{iz\left(\frac{9}{4}\left(z^2+2+\frac{1}{z^2}\right)
-\frac{4}{4}\left(z^2-2+\frac{1}{z^2}\right)\right)}\hfill$
($C$=unit circle)

$\ds\hspace{0.2in}=\frac{4}{i}\int_C\frac{zdz}{5z^4+26z^2+5}=
\frac{4}{i}\int_C\frac{zdz}{(5z^2+1)(z^2+5)}$

The integrand has simple poles inside $C$ at
$z=\pm\frac{i}{\sqrt5}$

Res$\ds\left(\frac{+i}{\sqrt5}\right)=
\left.\frac{z}{20z^3+52z}\right|_{\pm\frac{i}{\sqrt5}}=
\frac{1}{-4+52}=\frac{1}{48}$

So the integral is $\ds\frac{4}{i}2\pi
i\left(\frac{1}{48}+\frac{1}{48}\right)=\frac{\pi}{3}$

\end{itemize}

\end{document}
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