\documentclass[12pt]{amsart}

\newcommand{\B}{\mathcal{B}}
\newcommand{\C}{\mathcal{C}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\nc}{\mathbb{C}}
\newcommand{\nn}{\mathbb{N}}
\newcommand{\nq}{\mathbb{Q}}
\newcommand{\nr}{\mathbb{R}}
\newcommand{\nz}{\mathbb{Z}}

\theoremstyle{definition}
\newtheorem{prob}{Problem}

\newcommand{\abs}[1]{\lvert#1\rvert}

\begin{document}
\thispagestyle{empty}

\vspace*{-.7in}

\noindent
\textbf{Math 5616H} \hfill \textbf{Homework 5}

\noindent
\textbf{Posted: 2/21; Updated: 2/22; Due: Friday, 2/27/2015}\\
The problem set is due at the beginning of the class on
Friday. \smallskip

\noindent
\textbf{Reading}: 7.10, Chapter 8: pp. 172--174, 178--181,
182--185. \smallskip

\begin{prob}
  Using differentiation, determine the coefficients $a_n$, $n \ge 0$,
  of the power series whose sum is $(1-z)^{-2}$ for $\abs{z} < 1$.
\end{prob}

\begin{prob}
  Prove that, for any integer $k \ge 0$,
\[
\sum_{n=0}^\infty {n+k \choose k} z^n = \frac{1}{(1-z)^{k+1}}, \qquad
\abs{z} <1.
\]
\end{prob}

\begin{prob}
  Suppose that $f(z) = \sum_{n=1}^\infty a_n z^n$ has a radius of
  convergence $R > 0$, and suppose that $\abs{z_0} = r < R$. Define
\[
g(z) = \sum_{n=1}^\infty a_n (z-z_0)^n , \qquad \abs{z-z_0} < R - r.
\]
Prove that $g(z)$ is given by a convergent power series
\[
g(z) = \sum_{n=0}^\infty b_n z^n
\]
whose radius of convergence is at least $R-r$. \emph{Hint}: Use the
Binomial theorem to compute the coefficients $b_n$ and then use the
previous problem to estimate the $\limsup$ formula for the radius of
convergence of $g$ using the radius of convergence $R$ of $f$.
\end{prob}

\begin{prob}
  Determine the coefficients of the power series that defines a
  function with the following properties: $f''(z) = - f(z)$, $f(0) =
  1$, $f'(0) = 0$.
\end{prob}

\begin{prob}
  Show that $e^{z_1} = e^{z_2}$, where $e^z := E(z)$, if and only if
  $z_1 - z_2 = 2 \pi n i$ for some $n \in \nz$.
\end{prob}

\begin{prob}
Extend $C(x)$ and $S(x)$ from (46) in the textbook to complex $z$:
\[
\cos z = \frac{1}{2} (e^{iz} + e^{-iz}), \quad \sin z = \frac{1}{2i}
(e^{iz} - e^{-iz}).
\]
Is it true that for each $w \in \nc$ there is $z \in \nc$ such that
$\cos z = w$? Find all solutions when there are any.
\end{prob}

\begin{prob}
  Find ten other proofs of the Fundamental Theorem of Algebra. (No
  need to hand in) \end{prob}

\end{document}