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

{\bf Question}

\begin{itemize}
\item[a)]
Let $a$ be a fixed non-zero complex number.  Show that $a^i$ has
infinitely many values, and that they all lie on a straight line
through the origin in the complex plane.

\item[b)]
State Cauchy's integral formula expressing the nth derivative
$f^{(n)}(a)$ at a point $a$ in terms of a contour integral.  Your
statement should include conditions under which the formula holds.

Evaluate the following integrals, where $C$ denoted the unit
circle:

\begin{itemize}
\item[i)]
$\ds\int_C\frac{\cos zdz}{z^3}$

\item[ii)]
$\ds\int_C\frac{e^zdz}{4z^3-12z^2+9z-2}$
\end{itemize}

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$a^i=\exp(i\log a)=\exp(i(\ln|a|+i({\rm Arg} a+2n\pi))
\hspace{0.2in} n\in{\bf Z}$

$\ds\hspace{0.2in}=\exp(i\ln|a|)\exp(-{\rm Arg}a)\exp(-2n\pi)$

The first two factors are constants, and the third gives an
infinite sequence of distinct real numbers.  Hence we have
infinitely many different values, all on the line $\arg z=\ln|a|$.

$a^{\sqrt2}=\exp(\sqrt2\log a)=\exp(\sqrt2(\ln|a|+i({\rm Arg}
a+2n\pi))$

$\ds\hspace{0.3in}=\exp(\sqrt2\ln|a|)\exp(i\sqrt2-{\rm
Arg}a)\exp(i2\sqrt2n\pi)$

These all lie on the circle, centre 0, radius $|a|^{\sqrt2}$.
Since the members of the sequence $2\sqrt2n\pi$ are all different
${\rm mod} 2\pi$, since $\sqrt2$ is irrational, we have infinitely
many different values.

\item[b)]
If $f(z)$ is differentiable inside and on a closed contour $C$,
and if $a$ is inside $C$, then

$\ds f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}dz$

\begin{itemize}
\item[i)]
with $f(z)=\cos z, \,\,\, a=0, \,\,\, n=2,$

$\ds\int_C\frac{\cos z}{z^3}dz=\frac{2\pi i}{2!}f''(0)= \pi
i(-\cos 0)=-\pi i$

\item[ii)]
The denominator factorises as
$(z-2)(2z-1)^2=4(z-2)(z-\frac{1}{2})^2$

so with $\ds f(z)=\frac{e^z}{4(z-2)}, \,\,\, a=\frac{1}{2}, \,\,\,
n=1$

$\ds\int_C\frac{e^zdz}{4(z-2)(z-\frac{1}{2})^2}=\frac{2\pi
i}{1!}f'\left(\frac{1}{2}\right)$

Now $\ds
f'(z)=\frac{(z-2)e^z-e^z}{4(z-2)^2}=\frac{(z-3)e^z}{4(z-2)^2}$

So $\ds f'\left(\frac{1}{2}\right)=
\frac{-\frac{5}{2}e^{\frac{1}{2}}}{4\left(-\frac{3}{2}\right)^2}=
-\frac{5}{18}e^\frac{1}{2}. \hspace{0.5in}$ So
$\ds\int_C=-\frac{5}{9}\pi ie^\frac{1}{2}$

\end{itemize}

\end{itemize}

\end{document}
