Complex Numbers

Learning Objectives

In this section, you will:

• Add and subtract complex numbers.
• Multiply and divide complex numbers.
• Simplify powers of i𝑖

Figure 1

Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple.

In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that 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. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it.

Expressing Square Roots of Negative Numbers as Multiples of i𝑖

We know how to find the square root of any positive real number. In a similar way, we can find the square root of any 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 i𝑖 is defined as the square root of −1.−1.

−1−−−√=i−1=𝑖

So, using properties of radicals,

i2=(−1−−−√)2=−1𝑖2=(−1)2=−1

We can write the square root of any negative number as a multiple of i.𝑖. Consider the square root of −49.−49.

−49−−−−√===49⋅(−1)−−−−−−−√49−−√−1−−−√7i−49=49⋅(−1)=49−1=7𝑖

We use 7i7𝑖 and not −7i−7𝑖 because the principal root of 4949 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+bi𝑎+𝑏𝑖 where a𝑎 is the real part and b𝑏 is the imaginary part. For example, 5+2i5+2𝑖 is a complex number. So, too, is 3+4i3–√.3+4𝑖3.

Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers.

IMAGINARY AND COMPLEX NUMBERS

A complex number is a number of the form a+bi𝑎+𝑏𝑖 where

• a𝑎 is the real part of the complex number.
• b𝑏 is the imaginary part of the complex number.

If b=0,𝑏=0, then a+bi𝑎+𝑏𝑖 is a real number. If a=0𝑎=0 and b𝑏 is not equal to 0, the complex number is called a pure imaginary number. An imaginary number is an even root of a negative number.

HOW TO

Given an imaginary number, express it in the standard form of a complex number.

1. Write −a−−−√−𝑎 as a−−√−1−−−√.𝑎−1.
2. Express −1−−−√−1 as i.𝑖.
3. Write a−−√⋅i𝑎⋅𝑖 in simplest form.

EXAMPLE 1

Expressing an Imaginary Number in Standard Form

Express −9−−−√−9 in standard form.

Solution

−9−−−√==9–√−1−−−√3i−9=9−1=3𝑖

In standard form, this is 0+3i.0+3𝑖.

TRY IT #1

Express −24−−−−√−24 in standard form.

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),(𝑎,𝑏), where a𝑎 represents the coordinate for the horizontal axis and b𝑏 represents the coordinate for the vertical axis.

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

Figure 2

COMPLEX PLANE

In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown in Figure 3.

Figure 3

HOW TO

Given a complex number, represent its components on the complex plane.

1. Determine the real part and the imaginary part of the complex number.
2. Move along the horizontal axis to show the real part of the number.
3. Move parallel to the vertical axis to show the imaginary part of the number.
4. Plot the point.

EXAMPLE 2

Plotting a Complex Number on the Complex Plane

Plot the complex number 3−4i3−4𝑖 on the complex plane.

Solution

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

Figure 4

TRY IT #2

Plot the complex number −4−i−4−𝑖 on the complex plane.

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 then combine the imaginary parts.

COMPLEX NUMBERS: ADDITION AND SUBTRACTION

Adding complex numbers:

(a+bi)+(c+di)=(a+c)+(b+d)i(𝑎+𝑏𝑖)+(𝑐+𝑑𝑖)=(𝑎+𝑐)+(𝑏+𝑑)𝑖

Subtracting complex numbers:

(a+bi)−(c+di)=(a−c)+(b−d)i(𝑎+𝑏𝑖)−(𝑐+𝑑𝑖)=(𝑎−𝑐)+(𝑏−𝑑)𝑖

HOW TO

Given two complex numbers, find the sum or difference.

1. Identify the real and imaginary parts of each number.
2. Add or subtract the real parts.
3. Add or subtract the imaginary parts.

EXAMPLE 3

Adding and Subtracting Complex Numbers

Add or subtract as indicated.

1. ⓐ (3−4i)+(2+5i)(3−4𝑖)+(2+5𝑖)
2. ⓑ (−5+7i)−(−11+2i)(−5+7𝑖)−(−11+2𝑖)

Solution

We add the real parts and add the imaginary parts.

1. ⓐ
   (3−4i)+(2+5i)====3−4i+2+5i3+2+(−4i)+5i(3+2)+(−4+5)i5+i(3−4𝑖)+(2+5𝑖)=3−4𝑖+2+5𝑖=3+2+(−4𝑖)+5𝑖=(3+2)+(−4+5)𝑖=5+𝑖
2. ⓑ
   (−5+7i)−(−11+2i)====−5+7i+11−2i−5+11+7i−2i(−5+11)+(7−2)i6+5i(−5+7𝑖)−(−11+2𝑖)=−5+7𝑖+11−2𝑖=−5+11+7𝑖−2𝑖=(−5+11)+(7−2)𝑖=6+5𝑖

TRY IT #3

Subtract 2+5i2+5𝑖 from 3–4i.3–4𝑖.

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

Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, 3(6+2i)3(6+2𝑖) :

HOW TO

Given a complex number and a real number, multiply to find the product.

1. Use the distributive property.
2. Simplify.

EXAMPLE 4

Multiplying a Complex Number by a Real Number

Find the product 4(2+5i).4(2+5𝑖).

Solution

Distribute the 4.

4(2+5i)==(4⋅2)+(4⋅5i)8+20i4(2+5𝑖)=(4⋅2)+(4⋅5𝑖)=8+20𝑖

TRY IT #4

Find the product: 12(5−2i).12(5−2𝑖).

Multiplying Complex Numbers Together

Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term, i2,𝑖2, it equals −1.−1.

(a+bi)(c+di)===ac+adi+bci+bdi2ac+adi+bci−bd(ac−bd)+(ad+bc)ii2=−1Group real terms and imaginary terms.(𝑎+𝑏𝑖)(𝑐+𝑑𝑖)=𝑎𝑐+𝑎𝑑𝑖+𝑏𝑐𝑖+𝑏𝑑𝑖2=𝑎𝑐+𝑎𝑑𝑖+𝑏𝑐𝑖−𝑏𝑑𝑖2=−1=(𝑎𝑐−𝑏𝑑)+(𝑎𝑑+𝑏𝑐)𝑖Group real terms and imaginary terms.

HOW TO

Given two complex numbers, multiply to find the product.

1. Use the distributive property or the FOIL method.
2. Remember that i2=−1.𝑖2=−1.
3. Group together the real terms and the imaginary terms

EXAMPLE 5

Multiplying a Complex Number by a Complex Number

Multiply: (4+3i)(2−5i).(4+3𝑖)(2−5𝑖).

Solution

(4+3i)(2−5i)====4(2)−4(5i)+3i(2)−(3i)(5i)8−20i+6i−15(i2)(8+15)+(−20+6)i23−14i(4+3𝑖)(2−5𝑖)=4(2)−4(5𝑖)+3𝑖(2)−(3𝑖)(5𝑖)=8−20𝑖+6𝑖−15(𝑖2)=(8+15)+(−20+6)𝑖=23−14𝑖

TRY IT #5

Multiply: (3−4i)(2+3i).(3−4𝑖)(2+3𝑖).

Dividing Complex Numbers

Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form a+bi.𝑎+𝑏𝑖. 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+bi𝑎+𝑏𝑖 is a−bi.𝑎−𝑏𝑖. For example, the product of a+bi𝑎+𝑏𝑖 and a−bi𝑎−𝑏𝑖 is

(a+bi)(a−bi)==a2−abi+abi−b2i2a2+b2(𝑎+𝑏𝑖)(𝑎−𝑏𝑖)=𝑎2−𝑎𝑏𝑖+𝑎𝑏𝑖−𝑏2𝑖2=𝑎2+𝑏2

The result is a real number.

Note that complex conjugates have an opposite relationship: The complex conjugate of a+bi𝑎+𝑏𝑖 is a−bi,𝑎−𝑏𝑖, and the complex conjugate of a−bi𝑎−𝑏𝑖 is a+bi.𝑎+𝑏𝑖. 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+di𝑐+𝑑𝑖 by a+bi,𝑎+𝑏𝑖, where neither a𝑎 nor b𝑏 equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

c+dia+biwhere a≠0and b≠0𝑐+𝑑𝑖𝑎+𝑏𝑖where 𝑎≠0and 𝑏≠0

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

(c+di)(a+bi)⋅(a−bi)(a−bi)=(c+di)(a−bi)(a+bi)(a−bi)(𝑐+𝑑𝑖)(𝑎+𝑏𝑖)⋅(𝑎−𝑏𝑖)(𝑎−𝑏𝑖)=(𝑐+𝑑𝑖)(𝑎−𝑏𝑖)(𝑎+𝑏𝑖)(𝑎−𝑏𝑖)

Apply the distributive property.

=ca−cbi+adi−bdi2a2−abi+abi−b2i2=𝑐𝑎−𝑐𝑏𝑖+𝑎𝑑𝑖−𝑏𝑑𝑖2𝑎2−𝑎𝑏𝑖+𝑎𝑏𝑖−𝑏2𝑖2

Simplify, remembering that i2=−1.𝑖2=−1.

=ca−cbi+adi−bd(−1)a2−abi+abi−b2(−1)=(ca+bd)+(ad−cb)ia2+b2=𝑐𝑎−𝑐𝑏𝑖+𝑎𝑑𝑖−𝑏𝑑(−1)𝑎2−𝑎𝑏𝑖+𝑎𝑏𝑖−𝑏2(−1)=(𝑐𝑎+𝑏𝑑)+(𝑎𝑑−𝑐𝑏)𝑖𝑎2+𝑏2

THE COMPLEX CONJUGATE

The complex conjugate of a complex number a+bi𝑎+𝑏𝑖 is a−bi.𝑎−𝑏𝑖. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the result is a real number.

EXAMPLE 6

Finding Complex Conjugates

Find the complex conjugate of each number.

1. ⓐ 2+i5–√2+𝑖5
2. ⓑ −12i−12𝑖

Solution

1. ⓐThe number is already in the form a+bi.𝑎+𝑏𝑖. The complex conjugate is a−bi,𝑎−𝑏𝑖, or 2−i5–√.2−𝑖5.
2. ⓑWe can rewrite this number in the form a+bi𝑎+𝑏𝑖 as 0−12i.0−12𝑖. The complex conjugate is a−bi,𝑎−𝑏𝑖, or 0+12i.0+12𝑖. This can be written simply as 12i.12𝑖.

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.𝑖.

TRY IT #6

Find the complex conjugate of −3+4i.−3+4𝑖.

HOW TO

Given two complex numbers, divide one by the other.

1. Write the division problem as a fraction.
2. Determine the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.

EXAMPLE 7

Dividing Complex Numbers

Divide: (2+5i)(2+5𝑖) by (4−i).(4−𝑖).

Solution

We begin by writing the problem as a fraction.

(2+5i)(4−i)(2+5𝑖)(4−𝑖)

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

(2+5i)(4−i)⋅(4+i)(4+i)(2+5𝑖)(4−𝑖)⋅(4+𝑖)(4+𝑖)

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

(2+5i)(4−i)⋅(4+i)(4+i)====8+2i+20i+5i216+4i−4i−i28+2i+20i+5(−1)16+4i−4i−(−1)3+22i17317+2217iBecause i2=−1.Separate real and imaginary parts.(2+5𝑖)(4−𝑖)⋅(4+𝑖)(4+𝑖)=8+2𝑖+20𝑖+5𝑖216+4𝑖−4𝑖−𝑖2=8+2𝑖+20𝑖+5(−1)16+4𝑖−4𝑖−(−1)Because 𝑖2=−1.=3+22𝑖17=317+2217𝑖Separate real and imaginary parts.

Note that this expresses the quotient in standard form.

Simplifying Powers of i

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

i1=ii2=−1i3=i2⋅i=−1⋅i=−ii4=i3⋅i=−i⋅i=−i2=−(−1)=1i5=i4⋅i=1⋅i=i𝑖1=𝑖𝑖2=−1𝑖3=𝑖2⋅𝑖=−1⋅𝑖=−𝑖𝑖4=𝑖3⋅𝑖=−𝑖⋅𝑖=−𝑖2=−(−1)=1𝑖5=𝑖4⋅𝑖=1⋅𝑖=𝑖

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

i6=i5⋅i=i⋅i=i2=−1i7=i6⋅i=i2⋅i=i3=−ii8=i7⋅i=i3⋅i=i4=1i9=i8⋅i=i4⋅i=i5=i𝑖6=𝑖5⋅𝑖=𝑖⋅𝑖=𝑖2=−1𝑖7=𝑖6⋅𝑖=𝑖2⋅𝑖=𝑖3=−𝑖𝑖8=𝑖7⋅𝑖=𝑖3⋅𝑖=𝑖4=1𝑖9=𝑖8⋅𝑖=𝑖4⋅𝑖=𝑖5=𝑖

The cycle is repeated continuously: i,−1,−i,1,𝑖,−1,−𝑖,1, every four powers.

EXAMPLE 8

Simplifying Powers of i𝑖

Evaluate: i35.𝑖35.

Solution

Since i4=1,𝑖4=1, we can simplify the problem by factoring out as many factors of i4𝑖4 as possible. To do so, first determine how many times 4 goes into 35: 35=4⋅8+3.35=4⋅8+3.

i35=i4⋅8+3=i4⋅8⋅i3=(i4)8⋅i3=18⋅i3=i3=−i𝑖35=𝑖4⋅8+3=𝑖4⋅8⋅𝑖3=(𝑖4)8⋅𝑖3=18⋅𝑖3=𝑖3=−𝑖

TRY IT #7

Evaluate: i18𝑖18

Q&A

Can we write i35𝑖35 in other helpful ways?

As we saw in Example 8, we reduced i35𝑖35 to i3𝑖3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35𝑖35 may be more useful. Table 1 shows some other possible factorizations.

Factorization of i35𝑖35	i34⋅i𝑖34⋅𝑖	i33⋅i2𝑖33⋅𝑖2	i31⋅i4𝑖31⋅𝑖4	i19⋅i16𝑖19⋅𝑖16
Reduced form	(i2)17⋅i(𝑖2)17⋅𝑖	i33⋅(−1)𝑖33⋅(−1)	i31⋅1𝑖31⋅1	i19⋅(i4)4𝑖19⋅(𝑖4)4
Simplified form	(−1)17⋅i(−1)17⋅𝑖	−i33−𝑖33	i31𝑖31	i19𝑖19

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