\documentclass{amsart}

\pagestyle{empty}
\setlength{\parindent}{0pt}
\setlength{\textheight}{9in}
\setlength{\voffset}{-.75in}
\setlength{\textwidth}{6.5in}
\setlength{\hoffset}{-.75in}
%\everymath={\displaystyle}

\newcommand{\C}{\ensuremath{\mathbb{C}}}
\newcommand{\CC}{\ensuremath{\mathcal{C}}}
\newenvironment{problem}[1]{\leavevmode\llap{{#1~}}}{\vfill}

\renewcommand{\Re}{\mathrm{Re}}
\renewcommand{\Im}{\mathrm{Im}}

\begin{document}
\thispagestyle{empty}

\textbf{Math 425: Complex Variables}

\textbf{Quiz 3 Solutions}

\bigskip

\begin{problem}{1.}%
Prove that if $f(z)$ is entire and $\Re(f(z)) > 0$ for all $z$ then $f(z)$
is constant.

\bigskip

\emph{Solution.}
Let $w = u + iv$. Then $\big|\frac{w-1}{w+1}\big| < 1$ $\Leftrightarrow$
$|w-1| < |w+1|$ $\Leftrightarrow$ $u^2-2u+1+v^2 < u+2u+1+v^2$
$\Leftrightarrow$ $0<4u$.  Since $\Re(f(z))>0$, $f(z)$ never equals $-1$, and
so $g(z) = \frac{f(z)-1}{f(z)+1}$ is an entire function. By the above,
$|g(z)| < 1$ for all $z$, and so $g(z)$ is also bounded. Therefore, by
Liouville's Theorem, $g(z)$ is constant, $g(z) = c$. This implies $f(z) =
\frac{1+c}{1-c}$ is also constant.  \end{problem}

\begin{problem}{2.}%
Let $f$ be an entire function and let $\alpha$ be a real constant, $0 <
\alpha < 1$. Suppose $|f(z)| \le |z|^\alpha$ for all~$z$. Prove that $f$ is
identically zero.

\bigskip

\emph{Solution.} Let $z_0$ be any point in the complex plane. By Cauchy's
Inequality, $|f'(z_0)| \le \frac{M_R}{R}$ where $M_R = \max |f(z)|$ for
$|z-z_0| = R$. If $|z-z_0| = R$, then $|f(z)| \le |z|^\alpha = |z_0 +
(z-z_0)|^\alpha \le (|z_0| + |z-z_0|)^\alpha = (|z_0| + R)^\alpha$, so $M_R
\le (|z_0| + R)^\alpha$. Therefore, $|f'(z_0)| \le \frac{(|z_0| +
R)^\alpha}{R} = \Big[\frac{|z_0|}{R^{1/\alpha}} +
\frac{1}{R^{1/\alpha-1}}\Big]^\alpha \to 0$ as $R\to\infty$, because
$1/\alpha> 1$.  This shows that $f'(z_0) = 0$. But $z_0$ is arbitrary, so
$f'(z) = 0$ for all $z$, and $f(z)$ must be constant.  Since $|f(0)| \le
|0|^\alpha = 0$, that constant must be~$0$.

\bigskip

\emph{Alternate Solution.}
By Cauchy's Inequality, $|f^{(n)}(0)| \le \frac{n!M_R}{R^n}$ where $M_R =
\max|f(z)|$ for $|z|=R$. For $|z| = R$, $|f(z)| \le |z|^\alpha = R^\alpha$,
so $M_R \le R^\alpha$. Therefore, $|f^{(n)}(0)| \le
\frac{n!}{R^{n-\alpha}}\to 0$ as $R\to\infty$ and hence $f^{(n)}(0) = 0$ for
all $n\ge1$. This implies that in the power series for $f(z) =
\sum_{n=0}^\infty a_nz^n$ (which we already know converges for all $z$
because $f(z)$ is entire) the coefficients $a_n = \frac{f^{(n)}(0)}{n!} = 0$
for $n\ge1$. Note that $|f(0)| \le |0|^\alpha = 0$ so that $a_0 = 0$ as
well. Therefore, $f(z) = 0$ for all $z$.
\end{problem}

\begin{problem}{3.}%
Evaluate $\int_\CC \frac{e^{z}\sin(z)}{z^3}\,dz$ where $\CC$ is the circle
$|z| = 4$.

\bigskip

Let $f(z) = e^z\sin(z)$. Then $f'(z) = e^z(\sin(z)+\cos(z))$ and $f''(z) =
2e^z\cos(z)$. By the Cauchy Integral Formula, $\int_\CC
\frac{e^{z}\sin(z)}{z^3}\,dz = \frac{2\pi i}{2!} f''(0) = 2\pi i$.
\end{problem}

\begin{problem}{4.}%
Find the maximum of $\big|e^{-z(z+1)}\big|$  on the rectangle $-1 \le \Re(z)
\le 1$, $-1 \le \Im(z) \le 1$.

\bigskip

Since $f(z) = e^{-z(z+1)}$ is analytic for all~$z$, the Maximum Modulus
Principle says that the maximum of $|f(z)| = e^{-x^2-x+y^2}$ on the
rectangle occurs on the boundary of the rectangle. The boundary consists of
four line segments: $x = \pm1$ with $-1\le y\le 1$, or $y=\pm1$ with $-1\le
x\le 1$.

For $x=-1$, $g_1(y) = |f(-1 + iy)| = e^{y^2}$. To find the maximum of
$g_1(y)$ for $-1\le y\le 1$ we evaluate $g(y)$ at the critical point $y=0$
(where $g_1'(y) = 0$) and the endpoints: $g_1(0) = 1$ and $g_1(\pm1) = e$.
The maximum of $g_1(y)$ is~$e$.

On the segment $x=1$, $g_2(y) = |f(1+iy)| = e^{y^2-2}$. We evaluate
$g_2(y)$ at the critical point $y=0$ and the endpoints: $g_2(0) = e^{-2}$,
$g_2(\pm1) = e^{-1}$. So the maximum of $g_2(y)$ is~$e^{-1}$.

On the segments $y=\pm1$, $g_3(x) = |f(x\pm i)| = e^{-x^2-x+1}$. We
evaluate $g_3(x)$ at its critical point $x = -1/2$ ($g_3'(x) =
e^{-x^2-x+1}(-2x-1)=0 \Leftrightarrow x=-1/2$) and the endpoints:
$g_3(-1/2) = e^{5/4}$, $g_3(-1) = e$, $g_3(1) = e^{-1}$. The maximum of
$g_3(x)$ is $e^{5/4}$. Comparing all values we see that the maximum of
$|f(z)|$ is $e^{5/4}$ at the points $z = -1/2 \pm i$.
\end{problem}


\begin{problem}{5.}%
Derive the Taylor series for $\frac{1}{1+z}$ at the point $z = 1 + i$ and
determine its radius of convergence.
\end{problem}

\emph{Solution.} The radius of convergence is the distance of the center
$1+i$ to $-1$ since $z = -1$ is the only point where $f(z) = 1/(1+z)$ is
not analytic.  Thus, the radius  is $|1+i-(-1)| = |2+i| = \sqrt{5}$. We
obtain the series from the geometric series: Let $c = 1+i$. Then
  \begin{eqnarray*}
  \frac{1}{1+z} &=& \frac{1}{(1+c) + (z-c)}
  = \frac{1}{(1+c)}\cdot\frac{1}{1 - \big[-\frac{z-c}{1+c}\big]}
  =\frac{1}{(1+c)}\sum_{n=0}^\infty \Big[-\frac{z-c}{1+c}\Big]^n
  =\sum_{n=0}^\infty \frac{(-1)^n}{(1+c)^{n+1}}(z-c)^n\\
  &=&\sum_{n=0}^\infty \frac{(-1)^n}{(2+i)^{n+1}}(z-1-i)^n
  \end{eqnarray*}

\end{document}
