\section{Complex Numbers}
\begin{theorem}[Properties of Complex Numbers]
	\hfill
	\begin{enumerate}
		\item Closure under addition: $(a + bi) + (c + di) = (a + c) + (b + d)i$
		\item Closure under multiplication: $(a + bi)(c + di) = (ac) + (ad)i + (bc)i + (bd)i^2 = (ac - bd) + (ad + bc)i$
		\item Closure under division ($c + di \neq 0$): $\displaystyle\frac{(a + bi)}{(c + di)} = \frac{(a + bi)}{(c + di)}\frac{(c - di)}{(c - di)}=\frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \frac{(ac + bd)}{c^2 + d^2} + \frac{(bc - ad)}{c^2 + d^2}i$
		\item Complex conjugation: $(a + bi)(a - bi) = a^2 + b^2$
		\item Inverses: For every nonzero complex number $a + bi$ there is a complex number $c + di$ such that $(a + bi)(c + di)=1$. (That is, $(a + bi)^{-1}$ exists in $\C$.)
		\item Powers: For every complex number $a + bi = r(\cos\theta + i \sin \theta)$ and every positive integer $n$, we have $(a + bi)^n = [r(\cos \theta + i \sin \theta)]^n = r^n(\cos n \theta + i \sin n \theta)$.
		\item Radicals: For every complex number $a + bi = r(\cos \theta + i \sin \theta)$ and every positive integer $n$, we have $\displaystyle(a + bi)^{\frac{1}{n}} = [r(\cos \theta + i \sin \theta)]^{\frac{1}{n}} = r^{\frac{1}{n}}(\cos\frac{\theta}{n} + i \sin \frac{\theta}{n})$.
	\end{enumerate}
\end{theorem}