<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}

\begin{itemize}
\item[a)]
Let $C$ denote the spiral

$$\{z:z=e^{\theta^2+i\theta} \,\,\, 0\leq\theta\leq4\pi\}.$$

Evaluate $\ds\int_C\frac{dz}{z}$, giving a careful explanation of
your reasoning.


\item[b)]
Use Cauchy's Integral Formula to evaluate

$$\int_C\frac{e^z\sin^2\pi z}{(z-1)^3}dz$$

where $C$ is a simple closed contour having $z=1$ in its interior.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
On $C$, $z=e^{\theta^2}e^{i\theta}$

$z=e^{\theta^2}$ increase so $|z|$ increase with $\theta$.  So we
have a spiral


DIAGRAM


We need to use the logarithmic function, but we have to divide $C$
up into subsets on which $\arg z$ changes by less than $2\pi$,

on $\ds C_1 \hspace{0.5in} 0\leq\arg z\leq\frac{3\pi}{2}$

on $\ds C_2 \hspace{0.5in} \frac{3\pi}{2}\leq\arg z\leq3\pi$

on $\ds C_3 \hspace{0.5in} 3\pi\leq\arg z\leq4\pi$

on $\ds C_1 \hspace{0.5in}
\frac{d}{dz}\log_\frac{3\pi}{4}z=\frac{1}{z}$

on $\ds C_2 \hspace{0.5in}
\frac{d}{dz}\log_\frac{9\pi}{4}z=\frac{1}{z}$

on $\ds C_3 \hspace{0.5in}
\frac{d}{dz}\log_\frac{7\pi}{2}z=\frac{1}{z}$

so $\ds\int_C\frac{dz}{z}=
\int_{C_1}\frac{dz}{z}+\int_{C_2}\frac{dz}{z}+\int_{C_3}\frac{dz}{z}$

$\ds\hspace{0.5in}=[\log|z|+i\arg z]_{C_1}+[\log|z|+i\arg
z]_{C_2}+[\log|z|+i\arg z]_{C_3}$

Because $\arg z$ has been chosen continuously on $C$, this reduces
to

$[\log|z|+i\arg z]_C=\log e^{(4\pi)^2}-\log e^0+i4\pi=16\pi^2+4\pi
i$

${}$

Or this could be done directly from the definition of the integral
which is rather easier, as follows

$\ds\int_C\frac{dz}{z} \hspace{0.3in} z=e^{\theta^2+i\theta}
\hspace{0.3in} dz=(2\theta+i)e^{\theta^2+i\theta}$

so
$\ds\int_C\frac{dz}{z}=\int_0^{4\pi}2\theta+i=[\theta^2+i\theta]_0^{4\pi}=16\pi^2+4\pi
i$


\item[b)]

$\ds\int_C\frac{e^z\sin^2\pi z}{(z-1)^3}dz=\frac{2\pi
i}{2!}\frac{d^2}{dz^2}(e^z\sin^2\pi z)_{z=1}$

$\ds\frac{d}{dz}e^z\sin^2\pi z=e^z\sin^2\pi z+e^z2\sin\pi
z.\pi\cos\pi z$

$\ds\hspace{0.5in}=e^z\sin^2\pi z+e^z\pi\sin2\pi z$

$\ds\frac{d^2}{dz^2}e^z\sin^2\pi z=e^z\sin^2\pi z+e^z\pi\sin2\pi
z+e^z\pi\sin2\pi z+e^z4\pi^2\cos2\pi z$

At $z=1$

$\sin\pi z=0 \hspace{0.3in} \sin2\pi z=0 \hspace{0.3in} \cos2\pi
z=1$

so the integral is $\ds\frac{2\pi i}{2!}2\pi^2e=2\pi^3ei$

\end{itemize}


\end{document}
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