<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

4.5: Operations on Complex Numbers

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

  • Work with roots of quadratic equations by using the discriminant
  • Add and subtract complex numbers
  • Find the complex conjugate of a complex number
  • Multiply and divide complex numbers
  • Use your calculator to add, subtract, multiply, and divide complex numbers

Quadratic Formula

If ax2 + bx + c = 0


Recall that b2 - 4ac is called the discriminant.

If b2 - 4ac > 0 then there are two unequal real solutions.

If b2 - 4ac = 0 then there are two equal real solutions.

If b2 - 4ac < 0 then there are two unequal complex solutions.

Example: Solve x2 + 4x + 6 = 0.

Since b2 - 4ac = -8, the solutions will be complex


Sums and Differences of Complex Numbers

Recall that a complex number is given in the form a + bi where a and b are real numbers and i is the imaginary constant, i2 = -1.

When adding (or subtracting) two or more complex numbers the fastest method will be to add (or subtract) the real components to obtain the sum of the real numbers, and then separately add (or subtract) the imaginary coefficients to obtain the sum of the imaginary numbers or:

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

Example: Perform the Indicated Operation:

(6 + 3i) + (-5 + 2i) = (6 + -5) + (3 + 2) i = 1 + 5i

(3 – 2i) – (2 – 4i) = (3 - 2) + (-2 - ( -4)) i = 1 + 2i

(6) + (4 – 3i) = (6 + 4) + (0 + (-3)) i = 10 – 3i

Products and Quotients of Complex Numbers (conjugates)

Multiplying Complex Numbers:

When multiplying complex numbers in rectangular form, recall the method for multiplying two binomials (sometimes called FOIL): (m + n)(x + y) = mx + my + nx + ny. We use the same procedure for multiplying complex numbers:

(a + bi) + (c + di) = ac + adi + bci + bdi2

But, unlike the algebraic expression, the above expression contains the number, i.

Recall that i2 = -1, so bdi2 = bd(-1) = -bd and

adi + bci can be combined and then factored as (ad + bc)i. Thus we have the general result,

(a + bi) + (c + di) = (ac - bd) + (ad + bd)i

Example: Multiply: (6 + 3i)(2 - 3i)

(Combining like terms -9i2 reduces to -9(-1) or 9)

Example: Multiply: (5 - 7i)(5 + 7i)


Note: The product of the complex number (5 - 7i) and the complex number (5 + 7i) is a real number. When a complex number is multiplied by another complex number to produce a real number, the two complex numbers are called complex conjugates.

Example: Multiply i(6 - 2i)

But, the real part of a complex number is generally written first, so we can write this as

Dividing Complex Numbers

To divide two complex numbers is similar to dividing two irrational numbers. Recall that in that problem, the procedure was to find the irrational conjugate of the denominator and then multiply both the numerator and the denominator by this conjugate


First find the irrational conjugate of the denominator: , then multiply both the numerator and the denominator by this value:

this reduces to


In this case since you are interested in eliminating the complex numbers from the denominator, find the complex conjugate of the denominator and multiply BOTH the numerator AND the denominator by it.

You find the complex conjugate in the same way you found the conjugate of irrational numbers, change the sign of the imaginary part. For instance, the complex conjugate of 4 + 3i is 4 – 3i

A complex number multiplied by its complex conjugate will yield a real number. By recalling (a + b)(a - b) = a2 - b2 the complex conjugate can be found:

The conjugate of 4 + 3i is found by retaining the real part (4) and reversing only the sign of the imaginary part (that is, 3i becomes -3i)
(4 + 3i)(4 - 3i) = 16 - 12i + 12i - 9i2 Notice that -12i and 12i cancel. Also recall that i2 = -1
16 + 9 = 25 (4 + 3i)(4 – 3i) = 25

The product of this complex number and its conjugate is 25.

When multiplying complex numbers sometimes intuition about the nature of the product can mislead.

For example in (a + b)(a + b), where all of the terms are real numbers, no terms of each of the four products will cancel. Some of the terms may be combined. However in (1 + i)(1 - i), where some terms are real numbers and some terms are imaginary numbers, this is no longer true. Two of these terms cancel: the first product yields 1 while the last product yields i2 or -1, and those terms cancel!

Example: Find the quotient:

First, observe that the complex conjugate of the denominator is 4 – 3i

Multiply both the numerator and the denominator by 4 – 3i:

This number can also be written as or 0.6 – 0.24i

Applications, Technological Tools

A graphing calculator can perform operations with complex numbers. Press mode. Scroll down until: real a + bi re^ θi is seen, then select a + bi. Press Quit. Now the calculator is able to perform operations with complex numbers in a + bi form.

When the calculator is in complex number mode, be sure to use parenthesis to group the parts of the complex numbers. For example, enter 1 + 3i as (1 + 3i). Try doing some of the calculations from this section on your calculator to verify that complex mode works.

Lesson Summary

When adding and subtracting complex numbers add/subtract the real numbers and then add/subtract the imaginary numbers.

When multiplying complex numbers use the FOIL method of multiplication. Be sure to substitute i2 = -1 when appropriate.

When dividing complex numbers, write the problem as a fraction and then multiply both numerator and denominator by the conjugate of the denominator. If this process is done successfully, there will be only a real number in the denominator.

Points to Consider

Complex numbers- that is numbers that have a real and an imaginary part, also known as components, have characteristics similar to working with two unlike terms. However, they also have major differences such as not having as a final answer to a question, a denominator with an imaginary component or being sure to reduce i2 to -1.

Review Questions

Perform the indicated operations:

  1. a. (-8 - 2i) + (5 - 2i) b. (6 + i) - (5 - 2i) c. (-3i) - (2 - 3i) d. (3 + 2i)(5 - i) e. (0 + 4i)(3 - 1) f. (5 - 3i)2 g. h. (5 - 2i) ÷ (-2 + 3i)

Review Answers

  1. a. (3 - 4i) b. (1 + 3i) c. (-2 + 0i) d. (17 + 7i) e. (4 + 12i) f. (16 - 30i) g. h.

Image Attributions

Show Hide Details
Files can only be attached to the latest version of section
Help us create better content by rating and reviewing this modality.
Loading reviews...
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original

Original text