ex1.1.tex

\documentclass{hw}

\setlength{\parindent}{0pt}
\setlength{\parskip}{8pt}
%\usepackage[margin=.5in, paperwidth=8.5in, paperheight=11in]{geometry}

\usepackage{multicol}

\usepackage{mdwlist}

\usepackage{amssymb}
\usepackage{amsmath}

\usepackage{amsthm}
\newtheorem{thm}{Theorem}

\title{derp}
\author{derp}

\begin{document}

\problem{Express the following complex numbers in the form $a+ib$}
\begin{enumerate}
	\item $\displaystyle(2+3i)+(4+i)$
	\item $\displaystyle \frac{2+3i}{4+i}$
	\item $\displaystyle \frac{1}{i} + \frac{3}{1+i}$
\end{enumerate}
\solution
\part
\begin{equation*}4+2+(3+1)i = 6+4i\end{equation*}

\part 
\begin{equation*} \frac{2+3i}{4+i}\cdot\frac{4-i}{4-i} = \frac{8-2i+12i+3}{4^2-i^2} = \frac{11}{17} + \frac{10}{17}i \end{equation*}

\part
\noindent
\begin{align*}
	\frac{1}{i} = \frac{1}{i}\cdot\frac{i}{i} = -i\\
	\frac{3}{1+i}\cdot\frac{1-i}{1-i} = \frac{3-3i}{2} 
\end{align*}
Adding, 
\begin{equation*}
= \frac{3}{2} - \frac{3}{2}i -i =\frac{3}{2} - \frac{5}{2}i
\end{equation*}

\problem{Express the following complex numbers in the form $a+ib$}

\begin{enumerate}
	\item $\displaystyle (2+3i)(4+i)$
	\item $\displaystyle (8+6i)^2$
	\item $\displaystyle \left(1 + \frac{3}{1+i}\right)^2$
\end{enumerate}

\solution

\part

\begin{equation*}
	(2+3i)(4+i) = 8 + 2i + 12i + -3 = 5 + 14i
\end{equation*}

\part

\begin{equation*}
	(8+6i)^2 = 64+96i-36=28+96i
\end{equation*}

\part

From question 1, we have $\displaystyle \frac{3}{1+i} = \frac{3}{2} - \frac{3}{2}i$.

\begin{align*}
	\left(1 + \frac{3}{1+i}\right)^2 = \left(1+\frac{3-3i}{2}\right)^2 = 1 + 3-3i + \frac{(3-3i)^2}{4} \\
	=\frac{16-12i+(3-3i)^2}{4} = \frac{16-12i+9-18i-9}{4} = \frac{16+-30i}{4} = 4+\frac{15}{2}i
\end{align*}

\problem{Find the solutions to $z^2 = 3-4i$}
\solution
We want to find $z=x+iy$ ($x,y\in\mathbb{R}$) such that $z^2 = (x+iy)^2 = x^2 + 2xiy - y^2$.  Separating the imaginary and complex parts, we have:

\begin{eqnarray*}
	x^2-y^2 = 3\\
	2xy = -4 \Rightarrow x=-2/y
\end{eqnarray*}

Substituting, 
\begin{eqnarray*}
	\frac{4}{y^2} - y^2 -3 = 0 \\
	-y^4 -3y^2 +4 = 0 
\end{eqnarray*}
Factoring, we have $(y^2+4)(y^2-1) =0$, which gives $y^2 = -4$ or $y^2=1$. Since we require that $y$ be real, we can ignore the $-4$. Thus, we conclude that $y=1$ or $y=-1$ and:

\begin{eqnarray*}
	x = -2/y = -2/1 = -2 \\
	x = -2/-1 = 2
\end{eqnarray*}

It follows that $z=\pm(2-i)$

\problem{Find the solutions to the following equations:}
\begin{enumerate}
	\item $\displaystyle (z+1)^2 = 3+4i$
	\item $\displaystyle z^4-i = 0$
\end{enumerate}


\problem{Find the real and imaginary parts of the following, where $z=x+iy$.}

\solution

\part


\end{document}