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\begin{document}

{\bf Question}

\begin{itemize}
\item[i)]
State the Maximum Modulus Principle for a function $f(z)$ that is
analytic within a simple closed contour $\gamma$ and continuous on
$\gamma$.

Show that if $f(z)\not=0$ for all $z$ within and on $\gamma$, then
the minimum value of $|f(z)|$ cannot be achieved in the interior
of $\gamma$.

Hence find the maximum and minimum values of $|e^{z^2}|$ in the
disc $\{z|\,|z|\leq1\}$, and find where these values are attained.


\item[ii)]
Suppose that $f(z)$ and $g(z)$ are analytic in the disc
$\{z|\,|z|&lt;1\}$ and continuous on $|z|=1$.  Suppose that $f$ and
$g$ do not vanish for $|z|\leq1$, and that $|f(z)|=|g(z)|$
whenever $|z|=1$.  Prove that if $f(0)$ and $g(0)$ are positive
real numbers then $f(z)=g(z)$ whenever $|z|\leq1$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
First part bookwork.

${}$

Now $e^{z^2}\not=0$, so $|e^{z^2}|$ has max and min in $D$ on the
boundary.

Let $z=e^{i\theta}$, so $e^{z^2}=e^{2i\theta}$, and
$|e^{z^2}|=c^{\cos2\theta}$

so this is maximal where $\cos2\theta=+1$ i.e. $\theta=0, \pi$
i.e. max $e$.

and this is minimal where $\cos2\theta=-1$ i.e.
$\theta=\frac{\pi}{2}, \frac{3\pi}{2}$ i.e. min $e^{-1}$.


\item[ii)]
Let $\ds h(z)=\frac{f(z)}{g(z)}$ and $\ds k(z)=\frac{g(z)}{f(z)}$.

Both $h$ and $k$ are analytic in $|z|&lt;1$ since $f$ and $g$ are,
and are non-zero.  The max of $|h|$ and $|k|$ are both achieved on
the boundary, and in fact $|h|=|k|=1$ on the boundary.

So $\ds\frac{f(0)}{g(0)}\leq1$ and $\ds\frac{g(0)}{f(0)}\leq1$.
Thus $f(0)=g(0)=1$.

Thus $h$ has a local maximum at 0 and is therefore constant in $D$

i.e. $f(z)\equiv g(z)$ in $D$.

\end{itemize}

\end{document}
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