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

{\bf Question}

The function $f(z)$ is analytic inside and on a simple closed
curve $\gamma$.  Use Cauchy's theorem to prove Cauchy's integral
formula that for $b$ inside $\gamma$,

$$f(b)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)dz}{z-b}.$$

State a similar formula for the nth derivative $f^{(n)}(b)$.

Use these formulas to show that

\begin{itemize}
\item[i)]

$\ds\int_{\gamma}\frac{e^2z}{z^2}dz=4\pi i$

\item[ii)]
$\ds\int_{\gamma}\frac{\sin zdz}{9z^2+1}=\frac{2\pi
i}{3}\sinh\frac{1}{3}$.

\end{itemize}

where $\gamma$ is the circle centre 0 and radius 1.


\vspace{0.25in}

{\bf Answer}

Proof of Cauchy's Integral formula - bookwork.

\begin{itemize}
\item[i)]
Let $f(z)=e^{2z} \hspace{0.2in} f'(z)=2e^{2z}$

$\ds\int_{\gamma}\frac{f(z)}{(z-0)^2}dz=2\pi if'(0)=4\pi i$


\item[ii)]
$\ds\frac{1}{z^2+\frac{1}{9}}=
\frac{1}{\left(z+\frac{1}{3}i\right)\left(z-\frac{1}{3}\right)}=
\frac{3}{2i}\left(\frac{1}{z-\frac{1}{3}i}-
\frac{1}{z+\frac{1}{3}i}\right)$

$\ds\int_{\gamma}\frac{\sin z}{9z^2+1}dz=2\pi
i\frac{3}{2i}\frac{1}{9}\left\{\int_{\gamma}\frac{\sin
z}{z-\frac{1}{3}i}dz-\int_{\gamma}\frac{\sin
z}{z+\frac{1}{3}i}dz\right\}$

$\ds=\frac{\pi}{3}
\left(\sin\frac{1}{3}i-\sin\left(-\frac{1}{3}i\right)\right)=
\frac{2\pi i}{3}\sinh\frac{1}{3}$

\end{itemize}

\end{document}
