\documentclass[11pt]{article}
\setlength{\textheight}{8in}
\setlength{\topmargin}{0in}
\usepackage{graphicx, psfrag, amsmath, amsfonts}

\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\dis}{\displaystyle}
\newcommand{\bd}{\partial}

\begin{document}

\begin{center}{\bf \large Complex Numbers for Math 305}\end{center}

Complex numbers are used regularly in this course. It is assumed
you have worked with complex numbers before, but since they are not
used in the Calculus sequence, we review them here. The key idea is
to use a symbol $i$ to stand for a supposed $\sqrt{-1}$. The symbol $i$ is called
the imaginary unit. Using $i$ it has been proved that every polynomial
equation $p(z) = 0$ has a solution of the form $a + ib$, where $a$ and $b$ are
real numbers. 

The complex number field, denoted $\C$, is the set $\{a + bi\,|\,a, b \in \R\}$ subject to the rules
\[
(a + bi) + (c + di) = (a + c) + (b + d)i,
\]
and
\[
(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i.
\]
Subtraction is defined in the obvious way. Clearly, $\C$ contains the real
numbers $\R$ as a subset. Complex numbers of the form $bi$ are said to be
purely imaginary. Here are some examples.

\begin{itemize}
\item $(3 + 4i) - (7 - 2i) = -4 + 6i.$
\item $5i(2 + 3i) = -15 + 10i.$
\item $(2 - 6i) + 2i(1 + i) = (2 - 6i) + ( -2 + 2i) = - 4i.$
\item $i^{30} = i^{4\cdot7+2} = 1^7(i^2) = -1.$
\end{itemize}

Division is a bit subtle. The expression $w = 1/z$ means $zw = 1$.
Thus, if we want to find $1/(3+8i)$ we have to solve $(3+8i)(c+di) = 1$.
This gives $3c-8d = 1$ and $8c+3d = 0$. Solving this system of equations
gives $c = 3/(3^2 + 8^2)$ and $d = -8/(3^2 + 8^2)$. The general formula is
\[
\frac{1}{a+bi} = \frac{a}{a^2+b^2} + \frac{-b}{a^2+b^2}i.
\]


{\bf Example.}
\[
\frac{2+3i}{5-i} = \frac{(2+3i)(5+i)}{26} = \tfrac{7}{26} + \tfrac{17}{26}i.
\]

Initially, many viewed complex numbers as merely a mathematical
curiosity of little practical importance. But, in the 1800s it was discovered
that they had interesting geometrical properties which made them
useful in physics. We define the plane of complex numbers as having
the horizontal axis representing the real numbers and the vertical axis
representing the purely imaginary numbers. View the complex numbers
as vectors based at the origin. Notice that multiplication by $-1$
rotates a complex number by $180^o$ and that multiplication by $i$ rotates
a complex number by $90^o$ counterclockwise. Now the equation $i^2$ = $-1$
has a geometric meaning - it models a physically meaningful behavior.
Further, multiplication by $i^3 = -i$ rotates a complex number by $270^o$
counterclockwise or $90^o$ clockwise, and $i^4 = 1$ rotates a complex number
by $360^o$ which is to say multiplication by $1$ does nothing. See the
figure below for an example.

\begin{center}
\psfrag{a}{$5+3i$}
\psfrag{b}{$-3+5i$}
\psfrag{c}{$-5-3i$}
\psfrag{d}{$3-5i$}
\includegraphics[height=1.5in]{Figs/complex1.eps}
\end{center}

We introduce some terminology. The {\em complex conjugate} of a complex
complex number $z = x+yi$ is $\overline{z} = x-yi$. The {\em magnitude} of a complex
number $z = x + yi$ is $|z| = \sqrt{x^2 + y^2}$. Notice that $|z|^2 = z\overline{z}$. The
{\em argument} of a complex number $z = x + yi$ is the angle $\theta$ that the
vector $\left<x, y\right>$ makes with the positive real axis. The formula is
$\mbox{arg} (z) = \arctan(y/x)$, up to an additive factor of $\pi$ depending on the quadrant.
Notice that $1/z = \overline{z}/|z|^2$. Here are some examples.

\begin{itemize}
\item $|2 + 3i| = \sqrt{4 + 9} = \sqrt{13}$.
\item $(2 + 3i)\overline{(2 + 3i)} = (2 + 3i)(2 - 3i) = 4 - 6i + 6i - 9i^2 = 13$.
\item $\mbox{arg} (2 + 3i) = \arctan (3/2) \approx 0.98279\, \mbox{rad}$.
\item $\mbox{arg} (-2 + 3i) = \arctan (-3/2) +\pi \approx 2.15880\, \mbox{rad}$.
\end{itemize}

A complex number $z = x + yi$ can be expressed in polar form as
follows. Let $R = |z|$ and $\theta = \mbox{arg} (z)$. Then
\[
z = R(\cos \theta + i \sin \theta) = Re^{i\theta}.
\]
The first equality is just trigonometry. The second equality, called
{\em Euler's formula}, may seem bizarre at first, but it can be derived using
the Taylor series expansions of $\cos \theta$, $\sin \theta$ and $e^{i\theta}$.
This will be done
in class later. The polar form allows us to see the geometric meaning
of complex multiplication. Let $z = R_1e^{i\theta}$  and $w = R_2e^{i\phi}$.
Then
\[
zw = R_1R_2e^{i(\theta+\phi)}.
\]
In words, the magnitudes multiply while the arguments add.

A complex number $z$ is a {\em root of unity} if $z^n = 1$ for some positive
integer $n$. For $n = 2$ there are two roots of unity, $1$ and $-1$. For
$n = 4$ there are four roots of unity, 1, $i$, $-1$, and $-i$. What about
the roots of unity for $n = 3$? We can find these by observing that
$z^3 = 1$ can be written in polar form $z^3 = (Re^{i\theta})^3 = R^3e^{i3\theta} = 1$.
Set $R = 1$. Then we need for $3\theta$ to be a multiple of $2\pi$. This gives
$\theta = 0, \pm2\pi/3, \pm4\pi/3, \pm6\pi/3, \dots$. But, only three of
these are unique, 0, $2\pi/3$, and $4\pi/3$. These give 1, $\tfrac{-1}{2} +\tfrac{\sqrt{3}}{2}i$, and
 $\tfrac{-1}{2} -\tfrac{\sqrt{3}}{2}i$ as the three solutions to $z^3 = 1$.


For $n = 7$ there are seven roots of unity. The angles are $0$, $2\pi/7$, $4\pi/7$,
$6\pi/7$, $8\pi/7$, $10\pi/7$, $12\pi/7$. Thus, the seven solutions of $z^7 = 1$ are
$z =\cos (2k\pi/7) + i \sin (2k\pi/7)$ for $k = 0, 1, 2, 3, 4, 5, 6$. See the figures below.

\begin{center}
  \includegraphics[height=1.5in]{Figs/complex2.eps} \hspace{.3in}\includegraphics[height=1.5in]{Figs/complex3.eps}
\end{center}


{\bf Problem.} Solve $z^3 + 1 = 0$. This is similar to the roots of unity
problems, but we need to solve $z^3 = -1$. Again let $z = Re^{i\theta}$ be the
polar form. Set $R = 1$. Now we need for $3\theta$ to be an odd multiple of
$\pi$. Thus, $\theta = (2k + 1)\pi/3$. This gives $\pi/3$, $\pi$, and $5\pi/3$ as the possible
arguments. Thus the solution set for $z$ is
\[
\left\{ \tfrac{1}{2} +\tfrac{\sqrt{3}}{2}i, -1, \tfrac{1}{2} -\tfrac{\sqrt{3}}{2}i\right\}
\]

{\bf Problem.} Solve $z^4 +16 = 0$. Now we need $|z^4| = |-16|$. So, we let
$z = 2e^{i\theta}$.
We need for $4\theta = (2k + 1)\pi$. This gives $\pi/4$, $3\pi/4$, $5\pi/4$ and $7\pi/4$ as values for $\theta$.
Thus the solution set for $z$ is
\[
\left\{2 \cos ((2k + 1)\pi/4) + 2i \sin ((2k + 1)\pi/4)\,|\, k = 0, 1, 2, 3 \right\}
\]
\[
=\left\{ \sqrt{2} + i\sqrt{2}, -\sqrt{2} + i\sqrt{2}, -\sqrt{2} - i\sqrt{2}, \sqrt{2} - i\sqrt{2}\right\}.
\]

{\bf Exercise.} Find all the solutions of $(z^2 + 4)(z^5 -243)=0$. Plot them in the complex plane.
\end{document}

