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Appendix A
Complex Numbers

When you were very young and first learning about numbers, you probably began with the numbers we use for counting: \(\displaystyle 1, 2, 3,\) and so on. If you even had a concept of the number line at that time, it would have looked something like this:

In fact, until you grasped the concept of infinity, your number line may also have had a definite ending.

At some point, you learned about the concept of \(\displaystyle 0\text{.}\) During elementary school, negative numbers were introduced to you and your number line expanded to include all of the integers:

but it still had lots of gaps.

Once you learned about fractions, you began to fill in those gaps between the integers, and even though there were still technically some gaps, you probably thought of the number line as continuous. Eventually you learned about the real numbers, and your number line solidified:

Each of these steps required a real leap of the imagination. In many societies, mathematical reasoning stopped before one of these developments, and they never developed words or symbols for some or all of \(\displaystyle 0\text{,}\) negative numbers, fractions, or real numbers. These concepts were outside of their frame of reference, and would probably not have made sense to someone from one of these societies without a fair bit of explanation and justification. When you were young, you might well have been confused by the idea that someone could have less than no cookies. Understanding what a negative number means usually involves the introduction of the concept of debt, which is far less natural than counting things that you see. Similarly, the concept of fractional items only becomes natural if you have items that you want to divide up and the result of that division is not an integer. Real numbers typically only arise through either algebraic manipulation, or geometry, and even in those situations rational numbers are close enough for most practical purposes.

So maybe it's not so surprising that with complex numbers, we introduce a way to solve one of the few types of equation that have no solution in the real numbers: equations in which it is necessary to take the square root of a negative number. Is it really any more surprising to tell you that we can take the square root of a negative number after all, than it was to tell you that we could divide a single pie into seven equal pieces, obtaining a number that wasn't an integer? It does require us to once again expand our understanding of the number line, but we've done that before.

This, then, is the key concept of complex numbers: that \(\displaystyle \sqrt{-1}\) exists. Since it would be cumbersome to write \(\displaystyle \sqrt{-1}\) all the time, we give this quantity a symbol: \(\displaystyle i\text{.}\) Rather unfortunately, \(\displaystyle i\) is short for “imaginary”. While this is a natural term to use for the most basic number that isn't a “real” number, it does tend to reinforce many people's first impression on first learning about \(\displaystyle i\text{:}\) that these are numbers that don't really make any sense, or that have no practical value. In fact, complex numbers are extremely useful and no less natural than any of the other conceptual leaps we've been discussing.

Once we accept that there is no good reason for \(\displaystyle \sqrt{-1}\) not to exist, then why shouldn't \(\displaystyle \sqrt{-4}\) exist? In fact, it is an easy calculation that if \(\displaystyle i=\sqrt{-1}\) so that \(\displaystyle i^2=-1\text{,}\) then \(\displaystyle (2i)^2=4i^2=4(-1)=-4\text{,}\) so \(\displaystyle \sqrt{-4}=2i\text{.}\) Similarly, the square root of any real negative number is some real multiple of \(\displaystyle i\text{:}\) if \(\displaystyle x \in \mathbb{R^+}\) then \(\displaystyle \sqrt{-x}=\sqrt{x}\sqrt{-1}=\sqrt{x}i\text{.}\)

Actually, just as every equation of the form \(\displaystyle x^2=r\) has two real solutions when \(\displaystyle r\) is positive, it also has two solutions when \(\displaystyle r\) is negative. We see that \(\displaystyle (-i)^2=(-1)^2i^2=1(-1)=-1\) and similarly, if \(\displaystyle (ri)^2=x\) then \(\displaystyle (-ri)^2=x\) also.

At this point, in addition to the real number line we have introduced the imaginary number line: every possible real multiple of \(\displaystyle i\text{.}\) Since \(\displaystyle 0 \cdot i =0 \) is on both of these number lines, it is natural to draw the two number lines perpendicular to each other, intersecting at \(\displaystyle 0\text{:}\)

This image of the imaginary numbers led to a joke about an error message you might receive from your telephone service provider: “We're sorry; the number you have reached is imaginary. Please rotate your phone \(\displaystyle 90\) degrees and try your call again.”

It's pretty easy to see that if we take two numbers, each of which is on either the real or imaginary number line, and multiply them together, then we get another number that is on one of these lines. Also, if we add two real numbers the result is real. Similarly, if we add two imaginary numbers then the result is imaginary, since \(\displaystyle ai+bi=(a+b)i\text{.}\) But what if we add a real number to an imaginary number?

A number like \(\displaystyle 2+5i\) is not on either of our number lines. So again, we have to expand our concept of the numbers to include the whole of what we refer to as the complex plane. Any linear combination \(\displaystyle a+bi\) of a real number with an imaginary number, is a complex number. We can view this number as lying in position \(\displaystyle (a,b)\) of the plane, since its real part takes us \(\displaystyle a\) steps horizontally along the real number line, and its imaginary part takes us \(\displaystyle b\) steps vertically along the imaginary number line. Thus, any point in the plane corresponds to a complex number, and vice versa. In the image below we've inserted a small number of these points.

In high school, you should have learned how to use the quadratic formula to find the roots of quadratic equations. This is a common context in which complex numbers arise naturally. You were probably taught that for an equation of the form

\begin{equation*} ax^2+bx+c=0 \end{equation*}

the roots have the form

\begin{equation*} x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}. \end{equation*}

Very likely you learned that if \(\displaystyle b^2-4ac=0\) then there is only one (repeated) root (the graph of the quadratic equation just touches the \(\displaystyle x\)-axis at its maximum or minimum), and if \(\displaystyle b^2-4ac< 0\) then there are no roots (the graph of the equation is either entirely above, or entirely below, the \(\displaystyle x\)-axis). This is true insofar as the real roots are concerned, but when \(\displaystyle b^2-4ac< 0\) there are complex roots. When we allow complex numbers, an equation of degree \(\displaystyle d\) in \(\displaystyle x\) always has \(\displaystyle d\) roots (some of which may be repeated).

Now that we understand what complex numbers are and where they come from, we need to quickly cover the rules of basic arithmetic involving them.

Addition. We've already talked a bit about adding two complex numbers. If you're familiar with linear algebra and think of complex numbers as vectors in the plane, then adding them is just like vector addition. It's pretty straightforward even if you don't have that background. We have

\begin{equation*} (a+bi)+(c+di)=(a+c)+(b+d)i. \end{equation*}

We add the real parts of the summands to form the real part of the sum, and we add the complex parts of the summands to form the complex parts of the sum.

\begin{equation*} (1+\sqrt{2}i)+(-5+3i) = (1-5)+(3+\sqrt{2})i=-4+(3+\sqrt{2})i \end{equation*}

Subtraction. Subtraction is really similar to addition. We have

\begin{equation*} (a+bi)-(c+di)=(a-c)+(b-d)i. \end{equation*}

We take the difference of the real parts to form the real part of the difference, and we take the difference of the complex parts to form the complex parts of the difference.

\begin{equation*} (1+\sqrt{2}i)-(-5+3i) = (1-(-5))+(3-\sqrt{2})i=6+(3-\sqrt{2})i \end{equation*}

Multiplication. Multiplication is like multiplying polynomials: treat \(\displaystyle i\) as a variable, except that if we get an \(\displaystyle i^2\) we use the rule \(\displaystyle i^2=-1\) so that we always end up with something that looks like a real number plus an imaginary number:

\begin{equation*} (a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i. \end{equation*}
\begin{equation*} (1+\sqrt{2}i)(-5+3i) = -5+3i-5\sqrt{2}i+3\sqrt{2}i^2=(-5-3\sqrt{2})+(3-5\sqrt{2})i \end{equation*}

Before explaining division, we need to introduce complex conjugates.

Complex conjugation. Given a complex number \(\displaystyle a+bi\text{,}\) we refer to

\begin{equation*} a-bi \end{equation*}

as its complex conjugate. The value of this arises from the usefulness of differences of squares, and it will work in a similar manner to the technique of rationalising a denominator. If we take the product of a complex number with its complex conjugate, the result will be a difference of squares. Since the square of either a real number or a complex number is always real, this means that we get a real number! Here are the calculations:

\begin{equation*} (a+bi)(a-bi)=a^2-abi+abi-(bi)^2=a^2-b^2i^2=a^2-(-1)b^2=a^2+b^2. \end{equation*}

The complex conjugate of \(\displaystyle 1+\sqrt{2}i\) is \(\displaystyle 1-\sqrt{2}i\text{.}\) If we multiply these two values we obtain

\begin{equation*} (1+\sqrt{2}i)(1-\sqrt{2}i) =(1-(\sqrt{2}i)^2)=1+2=3. \end{equation*}

Division. If our denominator is a real nonzero number \(\displaystyle r\) then it is easy to perform division:

\begin{equation*} (a+bi)/r=(a/r)+(b/r)i. \end{equation*}

Otherwise, we use tricks (similar to those we might use if the denominator were irrational) to make the denominator easier to work with. In this case, we multiply both the numerator and the denominator by the complex conjugate of the denominator. This makes our denominator real, which we understand how to work with. Here's how we do this:

\begin{equation*} \frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(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. \end{equation*}
\begin{equation*} \frac{-5+3i}{1+\sqrt{2}i} = \frac{(-5+3i)(1-\sqrt{2}i)}{(1+\sqrt{2}i)(1-\sqrt{2}i)} =\frac{(-5+3\sqrt{2})+(3+5\sqrt{2})i}{1-(\sqrt{2}i)^2}=\left(-\frac{5}{3}+\sqrt{2}\right)+\left(1+\frac{5\sqrt{2}}{3}\right)i. \end{equation*}