Complex Numbers

In this section, I'll review basic properties of the complex numbers. The emphasis will be on the computational tools needed to use complex numbers in linear algebra.

A complex number is a number of the form $a + b i$ , where a and b are real numbers. a is called the real part and b is called the imaginary part; the notation is

$$\re (a + b i) = a, \quad \im (a + b i) = b.$$

i is a special symbol, sometimes called the complex or imaginary unit. It behaves somewhat like a variable, but it has a special multiplication property we'll see below.

The word "imaginary" is traditional, but not very good: It might lead you to believe that there's something "fake" about complex numbers, as opposed to the real numbers. In fact, the complex numbers are a number system like the real numbers or the integers, and there is nothing "fake" about them: I'll sketch a construction of the complex numbers using matrices below.

You can picture a complex number $a + b i$ as a point in the x-y-plane; we think of "a" as the x-coordinate and $b i$ (or b) as the y-coordinate.

$$\hbox{\epsfysize=1.5in \epsffile{complex-numbers-1.eps}}$$

Thus, the y-axis is an "i-axis".

The standard notation for the complex numbers is $\complex$ .

You add, subtract, and multiply complex numbers in the following ways:

$$(a + b i) + (c + d i) = (a + c) + (b + d)i,$$

$$(a + b i) - (c + d i) = (a - c) + (b - d)i,$$

$$(a + b i)(c + d i) = (a c - b d) + (a d + b c)i.$$

In the last equation, take $a = c = 0$ and $b = d = 1$ . The equation becomes

$$i \cdot i = -1.$$

This means that $i^2 = -1$ , so i is described as "the square root of -1". You might object that -1 doesn't have a square root. What is true is that no real number is a square root of -1, but the complex numbers are a different number system.

You can divide by (nonzero) complex numbers by "multiplying the top and bottom by the conjugate":

$$\dfrac{a + b i}{c + d i} = \dfrac{a + b i}{c + d i}\cdot \dfrac{c - d i}{c - d i} = \dfrac{(a c + bd) + (-a d + b c)i}{c^2 + d^2}.$$

With these operations, the set of complex numbers forms a field.

At this point, you might be a bit suspicious --- how can we just make up a bunch of symbols and operations and call the result "the complex numbers"? Here's a sketch which describes how you can "build" the complex numbers using matrices. The complex number $a + b i$ will correspond to the real $2 \times 2$ matrix

$$\left[\matrix{a & b \cr -b & a \cr}\right].$$

Setting $a = 0$ and $b = 1$ , we see that the complex number i should correspond to the matrix

$$\left[\matrix{0 & 1 \cr -1 & 0 \cr}\right].$$

The operations will be matrix addition and multiplication. Let's try them out:

$$\left[\matrix{a & b \cr -b & a \cr}\right] + \left[\matrix{c & d \cr -d & c \cr}\right] = \left[\matrix{a + c & b + d \cr -b - d & a + c \cr}\right],$$

$$\left[\matrix{a & b \cr -b & a \cr}\right] \left[\matrix{c & d \cr -d & c \cr}\right] = \left[\matrix{a c - b d & a d + b c \cr -a d - b c & a c - b d \cr}\right].$$

Notice that we're getting the same expressions we gave above for addition and multiplication of complex numbers. Thus, if you're comfortable with matrices and matrix arithmetic, you can think of "$a + b i$ " notation as shorthand for the matrix equivalents above. In an abstract algebra course, you'll probably see another construction of the complex numbers using quotient rings. In linear algebra, we'll just use the "$a + b i$ " form of complex numbers, as it is the simplest for computations.

To continue with our discussion of complex number arithmetic, I'll note here that the conjugate of complex number is obtained by flipping the sign of the imaginary part. The conjugate of $a + b i$ is denoted $\overline{a + b i}$ or sometimes $(a + b i)^*$ . Thus,

$$\overline{a + b i} = a - b i.$$

The norm of a complex number is

$$|a + b i| = \sqrt{a^2 + b^2}.$$

Note that

$$(a + b i)(\overline{a + b i}) = (a + b i)(a - b i) = a^2 + b^2 = |a + b i|^2.$$

When a complex number is written in the form $a + b i$ , it's said to be in rectangular form. There is another form for complex numbers that is useful: The polar form $r e^{i \theta}$ . In this form, r and $\theta$ have the same meanings that they do in polar coordinates.

DeMoivre's formula relates the polar and rectangular forms:

$$e^{i \theta} = \cos \theta + i \sin \theta.$$

This key result can be proven, for example, by expanding both sides in power series. Using this, we get

$$r e^{i \theta} = r \cos \theta + i \cdot r \sin \theta.$$

$$\hbox{\epsfysize=1.5in \epsffile{complex-numbers-2.eps}}$$

This relates the rectangular and polar forms of a complex number.

Note also that

$$|e^{i \theta}| = |\cos \theta + i \sin \theta| = \sqrt{(\cos \theta)^2 + (\sin \theta)^2} = 1.$$

Thus, $e^{i \theta}$ is a complex number of norm 1.

Example. Convert $3 + 4 i$ to polar form.

$$3 + 4 i = |3 + 4 i|\cdot \dfrac{3 + 4 i}{|3 + 4 i|} = 5\left(\dfrac{3}{5} + \dfrac{4}{5} i\right).$$

Let $\theta = \sin^{-1} \dfrac{4}{5}$ (or $\cos^{-1} \dfrac{3}{5}$ ). Then

$$3 + 4 i = 5(\cos \theta + i \sin \theta).\quad\halmos$$

In the examples that follow, we'll the polar and complex exponential forms of complex numbers to simplify algebraic computations, derive trig identities, and compute integrals.

Example. ( A trick with Demoivre's formula) Find $(1 + i \sqrt{3})^8$ .

It would be tedious to try to multiply this out. Instead, I'll try to write the expression in terms of $\cos \theta + i \sin \theta$ for a good choice of $\theta$ .

$$(1 + i \sqrt{3})^8 = 2^8 \cdot \left(\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)^8 = 256 \left(\cos \dfrac{\pi}{3} + i \sin \dfrac{\pi}{3}\right)^8 = 256 (e^{\pi i/3})^8 = 256 e^{8 \pi i/3} =$$

$$256 \left(\cos \dfrac{8 \pi}{3} + i \sin \dfrac{8 \pi}{3}\right) = 256 \left(\cos \dfrac{2 \pi}{3} + i \sin \dfrac{2 \pi}{3}\right) = 256 \left(-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}\right) = -128 + 128 i \sqrt{3}.\quad\halmos$$

Example. ( Proving trig identities) Prove the angle addition formulas:

$$\cos (a + b) = \cos a \cos b - \sin a \sin b.$$

$$\sin (a + b) = \sin a \cos b + \sin b \cos a.$$

I have

$$\eqalign{ e^{(a + b)i} & = e^{a i}e^{b i} \cr \cos (a + b) + i \sin (a + b) & = (\cos a + i \sin a)(\cos b + i \sin b) \cr \cos (a + b) + i \sin (a + b) & = (\cos a\cos b - \sin a \sin b) + i(\sin a \cos b + \sin b \cos a) \cr}$$

Equating real and imaginary parts on the left and right sides, I get

$$\cos (a + b) = \cos a \cos b - \sin a \sin b \quad\hbox{(real parts)}.$$

$$\sin (a + b) = \sin a \cos b + \sin b \cos a \quad\hbox{(imaginary parts)}.\quad\halmos$$

Example. ( Computing integrals) Compute $\displaystyle \int e^{2 x}\cos 3 x\,dx$ .

Note that $\cos 3 x = \re (e^{3 x i})$ . Thus,

$$\int e^{2 x}\cos 3 x\,dx = \int e^{2 x} \re (e^{3 x i})\,dx = \re \int e^{2 x}e^{3 x i}\,dx = \re \int e^{(2 + 3 i)x}\,dx = \re \dfrac{1}{2 + 3 i}e^{(2 + 3 i)x} + c =$$

$$\re \dfrac{2 - 3 i}{13}e^{2 x}(\cos 3 x + i \sin 3 x) + c = \dfrac{1}{13}e^{2 x} \re (2 - 3 i)(\cos 3 x + i \sin 3 x) + c =$$

$$\dfrac{1}{13}e^{2 x} \re \left((2 \cos 3 x + 3 \sin 3 x) + i(2 \sin 3 x - 3 \cos 3 x)\right) + c = \dfrac{1}{13}e^{2 x}(2 \cos 3 x + 3 \sin 3 x) + c.\quad\halmos$$

Suppose we have a polynomial of degree n with real coefficients:

$$a^n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0, \quad a_i \in \real.$$

The Fundamental Theorem of Algebra says that the polynomial has n roots which are real or complex numbers, some of which may be repeated. For example, the roots of $x^3 - 8$ are

$$2, \quad -1 + \sqrt{3}, \quad -1 - \sqrt{3}.$$

In general, finding the roots may require numerical approximation. You may see a proof of this theorem in advanced algebra or analysis courses; we'll take the result for granted.

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