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\assigntitle{16}{Complex Numbers (review)}
For the next couple weeks we will study the \term{complex} numbers,
$\C$. You've probably seen complex numbers before, so much of this
week's assignment might be review, but hopefully you'll learn some new
things as well. Next week, however, you'll learn some amazing things
about complex numbers which I guarantee you don't already know:
complex numbers, like vectors, can be represented both in rectangular
and polar form, and the polar form is totally sweet.\footnote{Oops, I
shouldn't have said anything. Now you probably won't be able to
sleep all week.}
\section{Imagine this}
\label{sec:i}
\begin{quote}
Suppose someone asked you to solve the equation \[ x^2 - 4 = 0. \]
``That's easy,'' you respond, ``$x = \pm 2$.''
``Okay,'' your mysterious interlocutor continues, ``how about \[ x^2 -
2 = 0\mbox{?''} \] ``No problem,'' you say,
``\begin{problem}\end{problem}''
The unknown interrogator presses, ``Aha, but what about \[ x^2 + 1 =
0\mbox{?''} \] ``Erm\dots'' you stammer. ``Everyone knows that
equation has \emph{no solutions}.'' You see, this is taking place 400
years ago.\footnote{Didn't you know that?}
``But isn't that weird?'' the aetherial inquisitor inquires. ``Some
polynomial equations have solutions, and some don't! It seems so
wrong\dots so\dots \emph{inelegant}.''
You concede that he has a good point, mumble some sort of excuse
involving a dentist, and leave as quickly as possible.
\end{quote}
\topic{history of imaginary numbers}
Funny as this sounds, it was the basic state of affairs in mathematics
until the 1700s or so. Gerolamo Cardano invented imaginary numbers in
the mid-1500s, but he didn't really understand them, and only used
them as tools for solving equations which actually had real solutions.
They didn't become widely accepted until Euler and Gauss used them in
the mid-1700s.
The interesting thing is where imaginary numbers got the name
``imaginary.'' Ren\'e Descartes\footnote{the same dude you learned
about in assignment \#14} was the first person to use that name for
them, in 1637---but he was using that name to \emph{make fun of them!}
Like many other mathematicians of the time, he dismissed them as
stupid, useless, and, well\dots \emph{imaginary}.
\topic{imaginary numbers aren't imaginary}
But ``imaginary'' numbers are actually no more imaginary than, say,
negative numbers. You cannot actually have \emph{negative three
rocks}. In some sense, ``negative three'' is an abstract, imaginary
concept that doesn't correspond to anything in the ``real world.''
But in another sense, it does correspond to things in the real
world---for example, if you owe someone three dollars, the situation
can be correctly and usefully modelled by the number ``negative
three.''
\topic{well, ok, they are, but so is everything else}
Fine, you say, but surely \emph{imaginary} numbers don't usefully
model anything in the real world. Ah, but they do! For example, the
voltage of the alternating current that is at this very moment
providing power to your computer can be represented by a complex
number. The quantum interactions of subatomic particles can also be
modeled using complex numbers. And there are probably many other such
things that I don't know about. The point is, once you get much
beyond numbers like $1$, $2$, $3$, \dots everything (fractions,
negative numbers, real numbers, humongously large integers\dots) is
pretty much imaginary. But that doesn't make them useless or
uninteresting.
\topic{the imaginary unit $i$}
Anyway, the key point to finding solutions for all those
``unsolvable'' equations, of course, is to introduce a new number
called $i$. This \term{imaginary unit} has the property that
\begin{equation}
\label{eq:im-unit}
i^2 = -1.
\end{equation}
\begin{problem}
Explain why there is no \emph{real number} which is a solution to
\pref{eq:im-unit}.
\end{problem}
In general:
\begin{defn}{imaginary numbers}
An \term{imaginary number} is any number of the form $ai$,
where $a$ is a real number, and $i$ is an \term{imaginary unit}
satisfying \pref{eq:im-unit}.
\end{defn}
It turns out that throwing $i$ in alongside the real numbers means
that now \emph{all} polynomial equations have solutions!
\begin{problem}
Prove that the square of every nonzero imaginary number is a
negative real number.
\end{problem}
\begin{problem}
\begin{subproblems}
\item You know what $i^2$ is. What is $i^3$?
\item What about $i^4$?
\item What is $i^{97}$?
\end{subproblems}
\end{problem}
\begin{problem}
Simplify: $\sqrt{-12}$.
\end{problem}
\section{Complex numbers}
\label{sec:complex}
If $3$ is a number, and $i$ is a number, what happens when we add
them? We can fix everything up by generalizing one more time, to
\term{complex numbers}. A complex number has a real part and an
imaginary part.
\begin{defn}{complex numbers}
A \term{complex number} is the sum of a real number and an imaginary
number; in other words, something of the form $a + bi$, where $a$
and $b$ are real numbers. The set of all complex numbers is
denoted $\C$ (\LaTeX: \verb|\C|).
\end{defn}
Again, ``complex'' is probably a bad name; there's nothing
particularly complex about the complex numbers. They're actually
pretty simple.
Keep in mind that the $b$ in $a + bi$ can be negative. For example,
$3 - 4i$ is a complex number; it is in the form $a + bi$ where $a = 3$
and $b = -4$.
\begin{problem}
Is $3$ a complex number? (\emph{Hint:} this is a trick question,
think carefully.)
\end{problem}
\begin{problem}
Solve for $x$:
\begin{subproblems}
\item $x^2 - 4x + 5 = 0$
\item $x^2 + x + 1 = 0$
\end{subproblems}
\end{problem}
\section{Complex arithmetic}
\label{sec:arithmetic}
\topic{basic rules for complex numbers}
Of course, it's no use having complex numbers if we can't do things
with them. Thankfully, manipulating complex numbers is very easy, if
you just keep in mind the following two rules:
\begin{enumerate}
\item $i^2 = -1$.
\item Combine terms with $i$ (imaginary parts) and terms without $i$
(real parts) separately.
\end{enumerate}
\topic{complex addition}
So, to add two complex numbers, just add the real parts, and add the
imaginary parts. For example, $(2 + 3i) + (4 - 6i) = 6 - 3i$.
\begin{problem}
Simplify:
\begin{subproblems}
\item $(19 - i) + (-2 + 6i)$
\item $(i - 4) + (5 - 2i)$
\item $(3 + i) - (4 + 6i)$
\item $(-5 + 2i) + 3(2 - i) - (4 - 4i)$
\end{subproblems}
\end{problem}
\topic{complex multiplication}
To multiply complex numbers, just expand out the multiplication,
remember that $i^2 = -1$, and collect like terms. For example:
\begin{align*}
(2 + 3i)(4 - i) &= (2)(4) + (2)(-i) + (3i)(4) + (3i)(-i) \\
&= 8 - 2i + 12i - 3i^2 \\
&= 8 - 2i + 12i + 3 \\
&= 11 + 10i
\end{align*}
\begin{problem}
Simplify:
\begin{subproblems}
\item $(2 + i)(3 + 2i)$
\item $(2 + i)^3$
\item $(6 + 2i)(6 - 2i)$
\end{subproblems}
\end{problem}
Dividing complex numbers is a bit trickier; to see how to do it, we'll
first need to explore the concept of \term{complex conjugates}.
\section{Complex conjugates}
\label{sec:conjugates}
\begin{defn}{conjugate}
The \term{conjugate} of the complex number $a + bi$ is $a - bi$.
\end{defn}
Of course, $b$ could be negative, so we can also say that the complex
conjugate of $a - bi$ is $a + bi$. The point is that to find the
conjugate, you switch the sign of the imaginary part.
\begin{problem}
What is the complex conjugate of $3$?
\end{problem}
\begin{problem}
What is the complex conjugate of $i + 4$?
\end{problem}
\begin{problem} \label{prob:prod-conjugate}
Prove that the product of any complex number with its conjugate is a
real number.
\end{problem}
\section{Complex division}
\label{sec:division}
\topic{complex division}
Now we are in a position to talk about division of complex numbers.
The idea is that we start with a fraction of the form \[ \frac{a+bi}{c
+ di}, \] and to simplify this we simply get rid of any imaginary
parts in the denominator. Of course, in \pref{prob:prod-conjugate}
you discovered a good way to do this: just multiply a complex number
by its conjugate and it becomes a real number, with no imaginary
parts. We can't just multiply the denominator by its
conjugate---but if we multiply the numerator \emph{and} denominator by
the conjugate of the denominator, then we're just multiplying by one
and everything is OK! Here's an example:
\begin{align*}
\frac{2 + 3i}{2 - 2i} &= \frac{2 + 3i}{2 - 2i} \cdot \frac{2 + 2i}{2
+ 2i} \\
&= \frac{4 + 6i + 4i - 6}{4 - 4i + 4i + 4} \\
&= \frac{-2 + 10i}{8} \\
&= \frac{-1 + 5i}{4}
\end{align*}
And now we're done, because we've simplified the division to a single
complex number (with $a = -1/4$ and $b = 5/4$).
\begin{problem}
Simplify:
\begin{subproblems}
\item $(-1 + 5i)/(1 + i)$
\item $(3 - 4i)/(2 - i)$
\item $5/i$
\end{subproblems}
\end{problem}
\end{document}