\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}}

\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 2.1  \\ Spring 2020 \hfill Due February 5}}}

\large

\vspace{1cm}


\begin{exercise}
Use de Moivre's formula and the binomial theorem to prove that 
\[
\cos(n \theta)  = \sum_{k=0}^{\lfloor n/2 \rfloor} {n \choose 2k} (-1)^k \cos^{n-2k} \theta \sin^{2k} \theta
\]
for all $n \in \N$.
\end{exercise}

\begin{exercise}  In this exercise we will give an alternate proof of the triangle inequality.

\begin{itemize}  

\item[\bf a.]  Prove that the triangle inequality is equivalent to the statement that $|1+r e^{i \theta}| \le 1+r$ for all $\theta \in \R$ and $r \ge 0$.  [{\em Suggestion:}  Use polar coordinates.]

\item[\bf b.]  Without using the triangle inequality, prove that $|1+r e^{i \theta}| \le 1+r$ for all $\theta \in \R$ and $r \ge 0$.  [{\em Suggestion:}  Consider $|1+re^{i \theta}|^2$.]

\end{itemize}

\end{exercise}

\begin{exercise}
Textbook exercise I.2.7.
\end{exercise}

\end{document}
