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

{\bf Question}

\begin{itemize}
\item[a)]
Find all values of $4^{4i}$, and show that they form a set of
point lying on a straight line in the complex plane.  Find the
limit point of this set.

\item[b)]
Let $C$ denote the semi-circle

$$\left\{e^{it}:-\frac{1}{2}\pi\leq t\leq\frac{1}{2}\pi\right\}.$$

Evaluate the contour integral

$$\int_C{\rm Log}zdz,$$

where Log$z$ denotes the principle logarithm of $z$.

\item[c)]
Let $K$ denote the circle, centre $1+i$ and radius 1.  Evaluate
the contour integral

$$\int_K\frac{{\rm Log}z}{z-(1+i)}dz,$$

giving your answer in the form $a+ib$, where $a$ and $b$ are real.

\end{itemize}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$a^b=\exp(b\log a)$

So $4^{4i}=\exp4i(\ln4+2n\pi i) \hspace{0.5in} n\in{\bf Z}$

$\hspace{0.3in}=\exp(i4\ln4)\exp(-8n\pi)$

These points all lie on the line $\arg z=4\ln4$, and have modulus

$\exp(-8n\pi)$.  They therefore have limit point zero.


\item[b)]
$\ds\int_C{\rm Log}zdz=\int_\frac{\pi}{2}^\frac{\pi}{2}({\rm
Log}|e^{it}|+i\arg e^{it})ie^{it}dt$

$\ds=\int_\frac{\pi}{2}^\frac{\pi}{2}-te^{it}dt=
\left[-t\frac{e^it}{i}\right]_\frac{\pi}{2}^\frac{\pi}{2}+
\int_\frac{\pi}{2}^\frac{\pi}{2}\frac{e^it}{i}dt$

$\ds=0+\left[-e^{it}\right]_\frac{\pi}{2}^\frac{\pi}{2}=-2i$


\item[c)]
Since $K$ does not meet the negative real axis, or contain 0,
Log$z$ is analytic inside and on $K$.  Thus

$\ds\int_K\frac{{\rm Log}z}{z-(1+i)}dz=2\pi i{\rm Log}(1+i)=2\pi
i\left(\ln\sqrt2+i\frac{\pi}{4}\right)=-\frac{\pi^2}{2}+\pi i\ln2$

\end{itemize}

\end{document}