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Appendix A Complex Numbers

Subsection Complex Numbers

Complex numbers are usually introduced as a tool to solve the quadratic equation \(x^2+1 = 0\text{.}\) However, that is not how complex numbers first came to light. The story actually involves solutions to the general cubic equation. The interested reader could consult Chapter 6 of William Dunham's excellent book Journey Through Genius. In this appendix we touch on the basics of complex numbers to provide enough context for the section on complex eigenvalues.

A complex number is defined by a pair of real numbers - the real part of the complex number and the imaginary part of the complex number.

Definition A.1.

A complex number is a number of the form

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

where \(a\) and \(b\) are real numbers and \(i^2 = -1\text{.}\)

The number \(a\) is the real part of the complex number and the number \(b\) is the imaginary part. We often write

\begin{equation*} z = a+bi \end{equation*}

for a complex number \(z\text{,}\) \(\text{ Re } (z)\) for the real part of \(z\) and \(\text{ Im } (z)\) for the imaginary part of \(z\text{.}\) That is, if \(z = a+bi\) with \(a\) and \(b\) real numbers, then \(\text{ Re } (z) = a\) and \(\text{ Im } (z) = b\text{.}\) We say that two complex numbers \(a+bi\) and \(c+di\) are equal if \(a=c\) and \(b=d\text{.}\)

There is an arithmetic of complex numbers that is determined by an addition and multiplication of complex numbers. Adding complex numbers is natural:

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

That is, to add two complex numbers we add their real parts together and their imaginary parts together.

Activity A.1.

Multiplication of complex numbers is is also done in a natural way.

(a)

By expanding the product as usual, treating \(i\) as we would any real number, and exploiting the fact that \(i^2 = -1\text{,}\) explain why we define the product of complex numbers \(a+bi\) and \(c+di\) as

\begin{equation*} (a+bi)(c+di) = (ac-bd) + (bc+ad)i\text{.} \end{equation*}

(b)

Use the definitions of addition and multiplication to write each of the sums or products as a complex number in the form \(a+bi\text{.}\)

(i)

\((2+3i) + (7-4i)\)

(ii)

\((4-2i)(3+i)\)

(iii)

\((2+i)i - (3+4i)\)

It isn't difficult to show that the set of complex numbers, which we denote by \(\C\text{,}\) satisfies many useful and familiar properties.

Activity A.2.

Show that \(\C\) has the same structure as \(\R\text{.}\) That is, show that for all \(u\text{,}\) \(w\text{,}\) and \(z\) in \(\C\text{,}\) the following properties are satisfied.

(a)

\(w+z \in \C\) and \(wz \in \C\)

(b)

\(w+z=z+w\) and \(wz=zw\)

(c)

\((w+z) + u = w + (z + u)\) and \((wz)u = w(zu)\)

(d)

There is an element \(0\) in \(\C\) such that \(z+ 0 = z\)

(e)

There is an element \(1\) in \(\C\) such that \((1)z = z\)

(f)

There is an element \(-z\) in \(\C\) such that \(z+(-z) = 0\)

(g)

If \(z \neq 0\text{,}\) there is an element \(\frac{1}{z}\) in \(\C\) such that \(z\left(\frac{1}{z}\right) = 1\)

(h)

\(u (w + z) = (u w) + (u z)\)

The result of Activity A.2 is that, just like \(\R\text{,}\) the set \(\C\) is a field. If we wanted to, we could define vector spaces over \(\C\) just like we did over \(\R\text{.}\) The same results hold.

Subsection Conjugates and Modulus

We can draw pictures of complex numbers in the plane. We let the \(x\)-axis be the real axis for a complex number and the \(y\)-axis the imaginary axis. That is, if \(z=a+bi\) we can think of \(z\) as a directed line segment from the origin to the point \((a,b)\text{,}\) where the terminal point of the segment is \(a\) units from the imaginary axis and \(b\) units from the real axis. For example, the complex numbers \(3+4i\) and \(-8+3i\) are shown in Figure A.2.

Figure A.2. Two complex numbers.

We can also think of the complex number \(z = a+bi\) as the vector \([a \ b]^{\tr}\text{.}\) In this way, the set \(\C\) is a two-dimensional vector space over \(\R\) with basis \(\{1, i\}\text{.}\) Each of these complex numbers has a length that we call the norm or modulus of the complex number. We denote the norm of a complex number \(a+bi\) as \(|a+bi|\text{.}\) The distance formula or the Pythagorean theorem show that

\begin{equation*} |a+bi| = \sqrt{a^2+b^2}\text{.} \end{equation*}

Note that

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

so the norm of the complex number \(a+bi\) can also be viewed as a square root of the product of \(a+bi\) with \(a-bi\text{.}\) The number \(a-bi\) is called the complex conjugate of \(a+bi\text{.}\) If we let \(z = a+bi\text{,}\) we denote the complex conjugate of \(z\) as \(\overline{z}\text{.}\) So \(\overline{a+bi} = a-bi\text{.}\)

Activity A.3.

Let \(w = 2+3i\) and \(z = -1+5i\text{.}\)

(a)

Find \(\overline{w}\) and \(\overline{z}\text{.}\)

(b)

Compute \(|w|\) and \(|z|\text{.}\)

(c)

Compute \(w\overline{w}\) and \(z \overline{z}\text{.}\)

(d)

Let \(z\) be an arbitrary complex number. There is a relationship between \(|z|\text{,}\) \(z\text{,}\) and \(\overline{z}\text{.}\) Find and verify this relationship.

(e)

What is \(\overline{z}\) if \(z \in \R\text{?}\)

Subsection Complex Vectors

A vector can have real and imaginary parts, too. For example, the vector \(\vv = \left[ \begin{array}{c} 1+i \\ 2-i \end{array} \right]\) can be written as

\begin{equation*} \vv = \left[ \begin{array}{c} 1+i \\ 2-i \end{array} \right] = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] + \left[ \begin{array}{r} i \\ -i \end{array} \right] = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] + i\left[ \begin{array}{r} 1 \\ -1 \end{array} \right]\text{.} \end{equation*}

The vector \(\text{ Re } (\vv) = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right]\) is the real part of \(\vv\) and the vector \(\text{ Im } (z) = \left[ \begin{array}{r} 1 \\ -1 \end{array} \right]\) is the imaginary part of \(\vv\text{.}\) In this way any vector \(\vv\) with complex entries can be written in the form

\begin{equation*} \vv = \vx + i \vy\text{,} \end{equation*}

where \(\vx = \text{ Re } (z)\) and \(\vy = \text{ Im } (z)\) are vectors with real entries. The conjugate of the vector \(\vv = \vx + i \vy\) is the vector \(\overline{\vv} = \vx - i \vy\text{.}\) We can do the same with matrices with complex entries.

There are several properties of complex conjugates that can be useful. Suppose \(r\) is a complex number, \(\vv\) is a vector with possibly complex entries, and \(A\) and \(B\) are matrices with possibly complex entries. Assume all products that are listed are defined. Then

  1. \(\displaystyle \overline{r\vv} = \overline{r} \ \overline{\vv}\)

  2. \(\displaystyle \overline{rA} = \overline{r} \ \overline{A}\)

  3. \(\displaystyle \overline{A\vv} = \overline{A} \ \overline{\vv}\)

  4. \(\displaystyle \overline{AB} = \overline{A} \ \overline{B}\)

These properties can be verified using complex arithmetic. For example, to verify the first property, let \(r = a+ib\) be a complex number and let \(\vv = \vx + i \vy\) be a complex vector. The operations on complex numbers give us

\begin{align*} \overline{r\vv} \amp = \overline{(a\vx-b\vy) + i(a\vy + b\vx)}\\ \amp = (a\vx-b\vy) - i(a\vy + b\vx)\\ \amp = (a-ib)(\vx - i\vy)\\ \amp = \overline{r} \overline{\vv}\text{.} \end{align*}

Verification of the remaining properties is left to the reader.