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\begin{document}

{\bf Question}

\begin{itemize}
\item[i)]
State Cauchy's integral formula (expressing $f(a)$ as a certian
integral around a closed curve surrounding $a$) paying particular
attention to the hypotheses under which the formula holds.  State
similar formulae for the nth derivatives $f^{(n)}(a)$.

Using these formulae, evaluate

\begin{itemize}
\item[a)]
$\ds\int_{|z|=1}\frac{\cos z}{z^3}dz$

\item[b)]
$\ds\int_{|z|=1}\frac{e^zdz}{4z^3-12z^2+9z-2}$
\end{itemize}


\item[ii)]
Use (i) to prove that if $f$ is analytic in a region $A$
containing a circle $\gamma$ with centre $a$ and radius $R$ and if
$|f(z)|\leq M$ on $\gamma$ then $|f^{(n)}(a)|\leq\frac{Mn!}{R^n}$.

Deduce Liouville's theorem that a function that is analytic and
bounded throughout {\bf C} is a constant function.

If $f(z)$ is analytic throughout {\bf C} and satisfies for all $z$
with $|z|>R$ an equality $|f(z)|\leq K|z|^{\frac{1}{2}}$, where
$K, \,\, R$ are positive real constants, prove that $f(z)$ is a
constant function.

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
If $f(z)$ is differentiable inside and on a closed contour $C$,
and if $a$ is inside $C$, then

$$f(a)=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-a}dz$$

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

\begin{itemize}
\item[a)]
with $f(z)=\cos z$

$\ds\int_{|z|=1}\frac{\cos z}{z^3}dz=\frac{2\pi i}{2!}f''(0)=\pi
i(-\cos0)=-\pi i$

\item[b)]
The denominator factorises as $(z-2)(2z-1)^2$,

so with $\ds g(z)=\frac{e^z}{4(z-2)}$

$\ds\int_{|z|=1}\frac{e^zdz}{(z-2)(2z-1)^2}=\frac{2\pi
i}{1!}g'\left(\frac{1}{2}\right)$

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

so $\ds g'\left(\frac{1}{2}\right)=
\frac{-\frac{5}{2}e^\frac{1}{2}}{4\left(\frac{3}{2}\right)^2}=
-\frac{5}{18}e^\frac{1}{2}$

so $\ds\int_{|z|=1}\frac{e^zdz}{(z-2)(2z-1)^2}=-\frac{5\pi
i}{9}e^\frac{1}{2}$

\end{itemize}


\item[ii)]
Using Cauchy's integral formula and the estimation lemma gives

$\ds|f^{(n)}(a)|=\left|\frac{n!}{2\pi
i}\int_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz\right|\leq
\frac{n!}{2\pi}\frac{M}{R^{n+1}}2\pi R=\frac{Mn!}{R^n}$

If $|f(z)|\leq M$ for all $z$ then this inequality holds for all
$R$, so $f^{(n)}(a)=0$ for all $n$.

Thus the Taylor series for $f$ consists just of the constant term
and so $f(z)=f(0)$ for all $z$.  This is Liouville's theorem.

Let $a$ be an arbitrary point of {\bf C}.  Let $C$ be a circle of
radius $r>|a|$, with $r>R$.

On $C$ $|f(z)|\leq Kr^\frac{1}{2}$

so $\ds|f'(a)|\leq\frac{Kr^\frac{1}{2}}{r}=
\frac{K}{r^\frac{1}{2}}\to0$ as $r\to\infty$

Thus $f'(a)=0$ for all $a$, so $f$ is constant.

\end{itemize}

\end{document}