Complex Numbers

The study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still have no solution to equations such as

x^{2} + 4 = 0

Our best guesses might be +2 or –2. But if we test +2 in this equation, it does not work. If we test –2, it does not work. If we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides solutions to problems such as these. In this section, we will explore this number system and how to work within it.

Expressing Square Roots of Negative Numbers as Multiples of i

We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number $i$ is defined as the square root of negative 1.

\sqrt{- 1} = i

So, using properties of radicals,

i^{2} = {(\sqrt{- 1})}^{2} = - 1

We can write the square root of any negative number as a multiple of $i .$ Consider the square root of –25.

\begin{array}{l} \begin{array}{l} \sqrt{- 25} = \sqrt{25 \cdot (- 1)} \end{array} \\ = \sqrt{25} \sqrt{- 1} \\ = 5 i \end{array}

We use $5 i$ and not $- 5 i$ because the principal root of $25$ is the positive root.

A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written $a + b i$ where $a$ is the real part and $b i$ is the imaginary part. For example, $5 + 2 i$ is a complex number. So, too, is $3 + 4 \sqrt{3} i .$

Showing the real and imaginary parts of 5 + 2i. In this complex number, 5 is the real part and 2i is the complex part.

Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.

Example 1

Expressing an Imaginary Number in Standard Form

Express $\sqrt{- 9}$ in standard form.

Solution

$\sqrt{- 9} = \sqrt{9} \sqrt{- 1} = 3 i$

In standard form, this is $0 + 3 i .$

Plotting a Complex Number on the Complex Plane

We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs $(a, b),$ where $a$ represents the coordinate for the horizontal axis and $b$ represents the coordinate for the vertical axis.

Let’s consider the number $−2 + 3 i .$ The real part of the complex number is $−2$ and the imaginary part is $3 i .$ We plot the ordered pair $(−2, 3)$ to represent the complex number $−2 + 3 i$ as shown in Figure 1.

Plot of a complex number, -2 + 3i. Note that the real part (-2) is plotted on the x-axis and the imaginary part (3i) is plotted on the y-axis. — Figure 1

Example 2

Plotting a Complex Number on the Complex Plane

Plot the complex number $3 - 4 i$ on the complex plane.

Solution

The real part of the complex number is $3,$ and the imaginary part is $−4 i .$ We plot the ordered pair $(3, −4)$ as shown in Figure 3.

Plot of a complex number, 3 - 4i. Note that the real part (3) is plotted on the x-axis and the imaginary part (-4i) is plotted on the y-axis. — Figure 3

Adding and Subtracting Complex Numbers

Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts.

Example 3

Adding Complex Numbers

Add $3 - 4 i$ and $2 + 5 i .$

Solution

We add the real parts and add the imaginary parts.

\begin{array}{l} (a + b i) + (c + d i) = (a + c) + (b + d) i \\ (3 - 4 i) + (2 + 5 i) = (3 + 2) + (- 4 + 5) i \\ = 5 + i \end{array}

Multiplying Complex Numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

Multiplying a Complex Number by a Real Number

Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

Showing how distribution works for complex numbers. For 3(6+2i), 3 is multiplied to both the real and imaginary parts. So we have (3)(6)+(3)(2i) = 18 + 6i.

Example 4

Multiplying a Complex Number by a Real Number

Find the product $4 (2 + 5 i) .$

Solution

Distribute the 4.

\begin{array}{l} 4 (2 + 5 i) = (4 \cdot 2) + (4 \cdot 5 i) \\ = 8 + 20 i \end{array}

Multiplying Complex Numbers Together

Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get

(a + b i) (c + d i) = a c + a d i + b c i + b d i^{2}

Because $i^{2} = - 1,$ we have

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

To simplify, we combine the real parts, and we combine the imaginary parts.

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

Example 5

Multiplying a Complex Number by a Complex Number

Multiply $(4 + 3 i) (2 - 5 i) .$

Solution

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

\begin{array}{l} (4 + 3 i) (2 - 5 i) = (4 \cdot 2 - 3 \cdot (- 5)) + (4 \cdot (- 5) + 3 \cdot 2) i \\ = (8 + 15) + (- 20 + 6) i \\ = 23 - 14 i \end{array}

Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of $a + b i$ is $a - b i .$

Note that complex conjugates have a reciprocal relationship: The complex conjugate of $a + b i$ is $a - b i,$ and the complex conjugate of $a - b i$ is $a + b i .$ Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose we want to divide $c + d i$ by $a + b i,$ where neither $a$ nor $b$ equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

\frac{c + d i}{a + b i} where a \neq 0 and b \neq 0

Multiply the numerator and denominator by the complex conjugate of the denominator.

\frac{(c + d i)}{(a + b i)} \cdot \frac{(a - b i)}{(a - b i)} = \frac{(c + d i) (a - b i)}{(a + b i) (a - b i)}

Apply the distributive property.

= \frac{c a - c b i + a d i - b d i^{2}}{a^{2} - a b i + a b i - b^{2} i^{2}}

Simplify, remembering that $i^{2} = −1.$

\begin{array}{l} = \frac{c a - c b i + a d i - b d (- 1)}{a^{2} - a b i + a b i - b^{2} (- 1)} \\ = \frac{(c a + b d) + (a d - c b) i}{a^{2} + b^{2}} \end{array}

Example 6

Finding Complex Conjugates

Find the complex conjugate of each number.

ⓐ $2 + i \sqrt{5}$
ⓑ $- \frac{1}{2} i$

Solution

ⓐ The number is already in the form $a + b i .$ The complex conjugate is $a - b i,$ or $2 - i \sqrt{5} .$
ⓑ We can rewrite this number in the form $a + b i$ as $0 - \frac{1}{2} i .$ The complex conjugate is $a - b i,$ or $0 + \frac{1}{2} i .$ This can be written simply as $\frac{1}{2} i .$

Analysis

Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by $i .$

Example 7

Dividing Complex Numbers

Divide $(2 + 5 i)$ by $(4 - i) .$

Solution

We begin by writing the problem as a fraction.

\frac{(2 + 5 i)}{(4 - i)}

Then we multiply the numerator and denominator by the complex conjugate of the denominator.

\frac{(2 + 5 i)}{(4 - i)} \cdot \frac{(4 + i)}{(4 + i)}

To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).

\begin{array}{l} \frac{(2 + 5 i)}{(4 - i)} \cdot \frac{(4 + i)}{(4 + i)} = \frac{8 + 2 i + 20 i + 5 i^{2}}{16 + 4 i - 4 i - i^{2}} \\ = \frac{8 + 2 i + 20 i + 5 (- 1)}{16 + 4 i - 4 i - (- 1)} & Because i^{2} = - 1 \\ = \frac{3 + 22 i}{17} \\ = \frac{3}{17} + \frac{22}{17} i & Separate real and imaginary parts . \end{array}

Note that this expresses the quotient in standard form.

Example 8

Substituting a Complex Number into a Polynomial Function

Let $f (x) = x^{2} - 5 x + 2.$ Evaluate $f (3 + i) .$

Solution

Substitute $x = 3 + i$ into the function $f (x) = x^{2} - 5 x + 2$ and simplify.

A step-by-step calculation showing the substitution of a complex number (3 + i) into a polynomial function, f(x) = x^2 - 5x + 2, and simplifying it to reach the final result of -5 + i, with explanations for each step.

Analysis

We write $f (3 + i) = −5 + i .$ Notice that the input is $3 + i$ and the output is $−5 + i .$

Example 9

Substituting an Imaginary Number in a Rational Function

Let $f (x) = \frac{2 + x}{x + 3} .$ Evaluate $f (10 i) .$

Solution

Substitute $x = 10 i$ and simplify.

\begin{array}{l} \frac{2 + 10 i}{10 i + 3} & Substitute 10 i for x . \\ \frac{2 + 10 i}{3 + 10 i} & Rewrite the denominator in standard form . \\ \frac{2 + 10 i}{3 + 10 i} \cdot \frac{3 - 10 i}{3 - 10 i} & Prepare to multiply the numerator and \\ denominator by the complex conjugate \\ of the denominator . \\ \frac{6 - 20 i + 30 i - 100 i^{2}}{9 - 30 i + 30 i - 100 i^{2}} & Multiply using the distributive property or the FOIL method . \\ \frac{6 - 20 i + 30 i - 100 (- 1)}{9 - 30 i + 30 i - 100 (- 1)} & Substitute –1 for i^{2} . \\ \frac{106 + 10 i}{109} & Simplify . \\ \frac{106}{109} + \frac{10}{109} i & Separate the real and imaginary parts . \end{array}

Simplifying Powers of i

The powers of $i$ are cyclic. Let’s look at what happens when we raise $i$ to increasing powers.

\begin{array}{l} i^{1} = i \\ i^{2} = - 1 \\ i^{3} = i^{2} \cdot i = - 1 \cdot i = - i \\ i^{4} = i^{3} \cdot i = - i \cdot i = - i^{2} = - (- 1) = 1 \\ i^{5} = i^{4} \cdot i = 1 \cdot i = i \end{array}

We can see that when we get to the fifth power of $i,$ it is equal to the first power. As we continue to multiply $i$ by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of $i .$

\begin{array}{l} i^{6} = i^{5} \cdot i = i \cdot i = i^{2} = - 1 \\ i^{7} = i^{6} \cdot i = i^{2} \cdot i = i^{3} = - i \\ i^{8} = i^{7} \cdot i = i^{3} \cdot i = i^{4} = 1 \\ i^{9} = i^{8} \cdot i = i^{4} \cdot i = i^{5} = i \end{array}

Example 10

Simplifying Powers of $i$

Evaluate $i^{35} .$

Solution

Since $i^{4} = 1,$ we can simplify the problem by factoring out as many factors of $i^{4}$ as possible. To do so, first determine how many times 4 goes into 35: $35 = 4 \cdot 8 + 3.$

i^{35} = i^{4 \cdot 8 + 3} = i^{4 \cdot 8} \cdot i^{3} = {(i^{4})}^{8} \cdot i^{3} = 1^{8} \cdot i^{3} = i^{3} = - i

QA Feature

Can we write $i^{35}$ in other helpful ways?

As we saw in Example 10, we reduced $i^{35}$ to $i^{3}$ by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of $i^{35}$ may be more useful. Table 1 shows some other possible factorizations.

Table 1 ..
Factorization of $i^{35}$	$i^{34} \cdot i$	$i^{33} \cdot i^{2}$	$i^{31} \cdot i^{4}$	$i^{19} \cdot i^{16}$
Reduced form	${(i^{2})}^{17} \cdot i$	$i^{33} \cdot (- 1)$	$i^{31} \cdot 1$	$i^{19} \cdot {(i^{4})}^{4}$
Simplified form	${(- 1)}^{17} \cdot i$	$- i^{33}$	$i^{31}$	$i^{19}$

Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.

Key Concepts

The square root of any negative number can be written as a multiple of $i .$ See Example 1.
To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See Example 2.
Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See Example 3.
Complex numbers can be multiplied and divided.
To multiply complex numbers, distribute just as with polynomials. See Example 4, Example 5, and Example 8.
To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See Example 6, Example 7, and Example 9.
The powers of $i$ are cyclic, repeating every fourth one. See Example 10.

Verbal

Exercise 1

Explain how to add complex numbers.

Solution

Add the real parts together and the imaginary parts together.

Exercise 2

What is the basic principle in multiplication of complex numbers?

Exercise 3

Give an example to show the product of two imaginary numbers is not always imaginary.

Solution

$i$ times $i$ equals –1, which is not imaginary. (answers vary)

Exercise 4

What is a characteristic of the plot of a real number in the complex plane?

Algebraic

For the following exercises, evaluate the algebraic expressions.

Exercise 5

$If f (x) = x^{2} + x - 4,$ evaluate $f (2 i) .$

Solution

$- 8 + 2 i$

Exercise 6

$If f (x) = x^{3} - 2,$ evaluate $f (i) .$

Exercise 7

$If f (x) = x^{2} + 3 x + 5,$ evaluate $f (2 + i) .$

Solution

$14 + 7 i$

Exercise 8

$If f (x) = 2 x^{2} + x - 3,$ evaluate $f (2 - 3 i) .$

Exercise 9

$If f (x) = \frac{x + 1}{2 - x},$ evaluate $f (5 i) .$

Solution

$- \frac{23}{29} + \frac{15}{29} i$

Exercise 10

$If f (x) = \frac{1 + 2 x}{x + 3},$ evaluate $f (4 i) .$

Graphical

For the following exercises, determine the number of real and nonreal solutions for each quadratic function shown.

Exercise 11

Graph of a parabola intersecting the real axis.

Solution

2 real and 0 nonreal

Exercise 12

Graph of a parabola not intersecting the real axis.

For the following exercises, plot the complex numbers on the complex plane.

Exercise 13

$1 - 2 i$

Solution

Exercise 14

$- 2 + 3 i$

Exercise 15

$i$

Solution

Exercise 16

$- 3 - 4 i$

Numeric

For the following exercises, perform the indicated operation and express the result as a simplified complex number.

Exercise 17

$(3 + 2 i) + (5 - 3 i)$

Solution

$8 - i$

Exercise 18

$(- 2 - 4 i) + (1 + 6 i)$

Exercise 19

$(- 5 + 3 i) - (6 - i)$

Solution

$- 11 + 4 i$

Exercise 20

$(2 - 3 i) - (3 + 2 i)$

Exercise 21

$(- 4 + 4 i) - (- 6 + 9 i)$

Solution

$2 - 5 i$

Exercise 22

$(2 + 3 i) (4 i)$

Exercise 23

$(5 - 2 i) (3 i)$

Solution

$6 + 15 i$

Exercise 24

$(6 - 2 i) (5)$

Exercise 25

$(- 2 + 4 i) (8)$

Solution

$- 16 + 32 i$

Exercise 26

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

Exercise 27

$(- 1 + 2 i) (- 2 + 3 i)$

Solution

$- 4 - 7 i$

Exercise 28

$(4 - 2 i) (4 + 2 i)$

Exercise 29

$(3 + 4 i) (3 - 4 i)$

Solution

Exercise 30

$\frac{3 + 4 i}{2}$

Exercise 31

$\frac{6 - 2 i}{3}$

Solution

$2 - \frac{2}{3} i$

Exercise 32

$\frac{- 5 + 3 i}{2 i}$

Exercise 33

$\frac{6 + 4 i}{i}$

Solution

$4 - 6 i$

Exercise 34

$\frac{2 - 3 i}{4 + 3 i}$

Exercise 35

$\frac{3 + 4 i}{2 - i}$

Solution

$\frac{2}{5} + \frac{11}{5} i$

Exercise 36

$\frac{2 + 3 i}{2 - 3 i}$

Exercise 37

$\sqrt{- 9} + 3 \sqrt{- 16}$

Solution

$15 i$

Exercise 38

$- \sqrt{- 4} - 4 \sqrt{- 25}$

Exercise 39

$\frac{2 + \sqrt{- 12}}{2}$

Solution

$1 + i \sqrt{3}$

Exercise 40

$\frac{4 + \sqrt{- 20}}{2}$

Exercise 41

$i^{8}$

Solution

$1$

Exercise 42

$i^{15}$

Exercise 43

$i^{22}$

Solution

$- 1$

Technology

For the following exercises, use a calculator to help answer the questions.

Exercise 44

Evaluate ${(1 + i)}^{k}$ for $k = 4, 8, and 12 .$ Predict the value if $k = 16.$

Exercise 45

Evaluate ${(1 - i)}^{k}$ for $k = 2, 6, and 10 .$ Predict the value if $k = 14.$

Solution

128i

Exercise 46

Evaluate $(1 + i)^{k} - (1 - i)^{k}$ for $k = 4, 8, and 12$ . Predict the value for $k = 16.$

Exercise 47

Show that a solution of $x^{6} + 1 = 0$ is $\frac{\sqrt{3}}{2} + \frac{1}{2} i .$

Solution

${(\frac{\sqrt{3}}{2} + \frac{1}{2} i)}^{6} = - 1$

Exercise 48

Show that a solution of $x^{8} - 1 = 0$ is $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i .$

Extensions

For the following exercises, evaluate the expressions, writing the result as a simplified complex number.

Exercise 49

$\frac{1}{i} + \frac{4}{i^{3}}$

Solution

$3 i$

Exercise 50

$\frac{1}{i^{11}} - \frac{1}{i^{21}}$

Exercise 51

$i^{7} (1 + i^{2})$

Solution

Exercise 52

$i^{−3} + 5 i^{7}$

Exercise 53

$\frac{(2 + i) (4 - 2 i)}{(1 + i)}$

Solution

5 – 5i

Exercise 54

$\frac{(1 + 3 i) (2 - 4 i)}{(1 + 2 i)}$

Exercise 55

$\frac{{(3 + i)}^{2}}{{(1 + 2 i)}^{2}}$

Solution

$- 2 i$

Exercise 56

$\frac{3 + 2 i}{2 + i} + (4 + 3 i)$

Exercise 57

$\frac{4 + i}{i} + \frac{3 - 4 i}{1 - i}$

Solution

$\frac{9}{2} - \frac{9}{2} i$

Exercise 58

$\frac{3 + 2 i}{1 + 2 i} - \frac{2 - 3 i}{3 + i}$

complex conjugate

the complex number in which the sign of the imaginary part is changed and the real part of the number is left unchanged; when added to or multiplied by the original complex number, the result is a real number

complex number

the sum of a real number and an imaginary number, written in the standard form

a + b i,

where

a

is the real part, and

b i

is the imaginary part

complex plane

a coordinate system in which the horizontal axis is used to represent the real part of a complex number and the vertical axis is used to represent the imaginary part of a complex number

imaginary number

a number in the form

b i

where

i = \sqrt{- 1}