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\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
State Liouville's Theorem, and use it to prove the Fundamental
Theorem of Algebra.

\item[b)]
Locate the zeros and singularities of the function

$$\frac{(z^2-4)\cos\left(\frac{1}{z}\right)}{z^2+z-6}$$

Classify the singularities, and determine the behaviour of the
function at infinity.

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Liouville's Theorem states that if $f(z)$ is analytic for all $z$,
and is bounded then $f(z)$ is constant.

Let $P(z)=z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_n \hspace{0.2in}
n\geq1$ be a polynomial (with complex coefficients).  Suppose
$P(z)\not=0$, for all $z\in{\bf C}$.

Let $f(z)=\frac{1}{P(z)}$, and so $f(z)$ is analytic for all
$z\in{\bf C}$.

Now $\frac{P(z)}{z^n}\rightarrow 1$ as $|z|\to\infty$.  Thus
$\exists R, \,\,\, |z|&gt;R \Rightarrow |f(z)|&lt;1$.  Now since $f(z)$
is continuous, it is bounded in $|z|\leq R$.  Thus $f(z)$ is
bounded and so constant. So $f(z)\equiv
f(0)=\frac{1}{P(o)}\not=0$.

This contradicts $f(z)\rightarrow0$ as $|z|\to\infty$.  Hence
$P(z)$ must have a zero.


\item[b)]
$\ds
f(z)=\frac{(z+2)(z-2)\cos\left(\frac{1}{z}\right)}{(z-2)(z+3)}$

So we have

\begin{itemize}
\item[i)]
a removable singularity at $z=2$.

\item[ii)]
a simple pole at $z=-3$.

\item[iii)]
an essential singularity at $z=0$.

\item[iv)]
a zero at $z=-2$.

\item[v)]
zeros where $\frac{1}{z}=(2n+1)\frac{\pi}{2}, \hspace{0.5in}
n\in{\bf Z}$
\end{itemize}

$\ds f\left(\frac{1}{z}\right)=
\frac{\left(\frac{1}{z}+2\right)\left(\frac{1}{z}-2\right)\cos
z}{\left(\frac{1}{z}-2\right)\left(\frac{1}{z}+3\right)}\rightarrow
1$ as $z\to0$

Hence $f$ is analytic at infinity.

\end{itemize}

\end{document}
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