\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Throughout this question the branch of $\log z$ whose imaginary
part $v$ satisfies $-\pi<v\leq\pi$ is used.

\begin{itemize}
\item[i)]
Let $\gamma$ be the circle $|z|=R$, where $R>1$.  Stating
carefully any inequalities concerning integrals you use prove that

$$\left|\int_{\gamma}\frac{\log
zdz}{z^2}\right|\leq2\pi\left(\frac{\pi+\log R}{R}\right)$$


\item[ii)]
Evaluate $\ds\int_{\delta}\log zdz$, where $\delta$ is the upper
half of the unit circle from $z=1$ to $z=-1$.


\item[iii)]
Let $n$ denote the circle with centre $\ds\frac{1+\sqrt3i}{2}$ and
radius $\frac{1}{2}$.  Use the Cauchy Integral Formula to evaluate

$$\int_n\frac{\log zdz}
{z-\left(\frac{1}{2}+\frac{\sqrt3i}{2}\right)}$$

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
On $\gamma$, $|\log z|=|\log R+i\theta|\leq\log R+\pi$

since $|z|=R$, $\ds\left|\frac{\log z}{z^2}\right|\leq\frac{\log
R+\pi}{R^2}$

$l(\gamma)=2\pi R$, so $\ds\left|\int_{\gamma}\frac{\log
z}{z^2}dz\right|\leq2\pi\left(\frac{\log R+\pi}{R}\right)$


\item[ii)]
On $\delta$, $z=e^{i\theta}$ and $\log z=i\theta$,
$0\leq\theta\leq\pi$

So $\ds\int_{\delta}\log zdz=\int_0^\pi i\theta
ie^{i\theta}d\theta=2-\pi i$   (by parts)


\item[iii)]
$n$ lies in the first quadrant so $\log z$ is analytic inside and
on $n$

So $\ds\int_n\frac{\log
z}{z-\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)}=
\log\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)=
\log1+i\frac{\pi}{3}=i\frac{\pi}{3}$

\end{itemize}


\end{document}
