\documentclass{article}
\usepackage{amsmath, amssymb,verbatim, amsthm, graphicx}

%exercise environment
\newcounter{exercise}
\setcounter{exercise}{0}
\newenvironment{exercise}{\addtocounter{exercise}{1} \noindent{\bf{Exercise \theexercise.}}}{\vspace{.5cm}}

\newtheorem{thm}{Theorem}

\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\C}{\mathbb{C}}

\textwidth 6.5in
\textheight 9in
\oddsidemargin 0in
\evensidemargin 0in
\headsep -0.5in

\setlength{\parskip}{0.25cm}

\pagenumbering{gobble}

\begin{document}

\begin{center}
\includegraphics[width=0.5in]{../ltr-logo.jpg}
\end{center}

\noindent{\Large{\textsc{Complex Variables\hfill Assignment 9.2  \\ Spring 2020 \hfill Due April 8}}}

\large

\vspace{1cm}

\begin{exercise}
In this exercise you will prove the following strengthening of Cauchy's theorem for rectangles.

\begin{thm}
Let $\Omega \subset \C$ be a domain and $z_0 \in \Omega$.  Suppose $f: \Omega \setminus \{ z_0 \} \to \C$ is analytic and satisfies $\lim_{z \to z_0} (z-z_0)f(z) = 0$.  If $R$ is a closed rectangular region in $\Omega$ and $z_0 \not \in \partial R$, then
\[
\int_{\partial R} f(z) \, dz = 0.
\]
\end{thm}

\begin{itemize}

\item[\bf a.]  Explain why it suffices to assume $z_0$ is in the interior of $R$.  [{\em Suggestion:}  Appeal to the original version of Cauchy's theorem for rectangles.]

\item[\bf b.]  Subdivide $R$ as shown below.  Let $R_0$ denote the shaded rectangle.  Explain why
\[
\int_{\partial R} f(z) \, dz = \int_{\partial R_0} f(z) \, dz.
\]
[{\em Suggestion:}  See the previous suggestion.]

\begin{center}
\includegraphics[width=2in]{rectangle.eps}
\end{center}

\item[\bf c.]  Let $\epsilon > 0$ and choose $\delta > 0$ so that $0 < |z-z_0| < \delta$ implies $|(z-z_0) f(z)| < \epsilon$.  Show that $|f(z)| < 2 \epsilon/\delta$ along $\partial R_0$.

\item[\bf d.]  Use an $ML$ estimate to show that
\[
\left| \int_{\partial R_0} f(z) \, dz \right| < 8 \epsilon,
\]
and use this to finish the proof.

\end{itemize}


\end{exercise}


\end{document}
