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Copyright © Cengage Learning. All rights reserved. 8.5 Complex Numbers Copyright © Cengage Learning. All rights reserved.

Objectives Identify complex numbers. Perform arithmetic operations with complex numbers. Find complex solutions to equations.

Definition of Imaginary and Complex Numbers

Definition of Imaginary and Complex Numbers We have noted that the square root of a negative number is not a real number. Today, something like is considered a nonreal number and is called an imaginary number. Because these numbers were not believed to really exist, they were called imaginary numbers. Later they were proven to exist, and they have been shown to be applicable in different fields of mathematics and science.

Definition of Imaginary and Complex Numbers Although imaginary numbers were proven to exist, the name had stuck by then, so we must remember that the name imaginary does not mean that these numbers do not exist. In mathematics, the number is the imaginary unit and is usually represented by the letter i. Using the letter i, we can represent imaginary numbers without showing a negative under a square root.

Definition of Imaginary and Complex Numbers Imaginary numbers can be combined with the real numbers into what are called complex numbers. Any number that can be written in the form a + bi is a complex number; a is considered the real part of a complex number, and b is considered the imaginary part.

Definition of Imaginary and Complex Numbers All real numbers are considered complex numbers whose imaginary part is equal to zero. All imaginary numbers are also considered complex numbers whose real part is equal to zero.

Definition of Imaginary and Complex Numbers

Definition of Imaginary and Complex Numbers Complex numbers are used in many areas of mathematics and many physics and engineering fields. Electronics uses complex numbers to work with voltage calculations. The shapes of airplane wings are developed and studied by using complex numbers. Many areas of algebra, such as fractals and chaos theory, can be studied by working in the complex number system.

Definition of Imaginary and Complex Numbers

Example 2 – Name the parts of complex numbers For each complex number, name the real part and the imaginary part. a. 5 + 4i b. 23 + 7i c. 5i d. 9

Example 2 – Solution a. 5 + 4i; Real part = 5; imaginary part = 4. b. –3 + 7i; Real part = –3; imaginary part = 7. c. In standard complex number form, 5i = 0 + 5i. Real part = 0; imaginary part = 5. d. In standard complex number form, 9 = 9 + 0i. Real part = 9; imaginary part = 0.

Operations with Complex Numbers

Operations with Complex Numbers Complex numbers can be added and subtracted easily by adding or subtracting the real parts together and then adding or subtracting the imaginary parts together. This is similar to combining like terms with variables.

Example 3 – Add or subtract complex numbers Add or subtract the following complex numbers. a. (2 + 8i) + (6 + 7i) b. (5 – 4i) + (7 + 6i) c. (6 + 3i) – (4 + 8i) d. (2.5 + 3.8i) – (4.6 – 7.2i)

Example 3 – Solution a. (2 + 8i) + (6 + 7i) = 2 + 6 + 8i + 7i b. 12 + 2i c. (6 + 3i) – (4 + 8i) = 6 + 3i – 4 – 8i = (6 – 4) + (3i – 8i) = 2 – 5i Add the real parts. Add the imaginary parts. Distribute the negative sign. Subtract the real parts. Subtract the imaginary parts.

Example 3 – Solution d. (2.5 + 3.8i) – (4.6 – 7.2i) cont’d d. (2.5 + 3.8i) – (4.6 – 7.2i) = (2.5 + 3.8i – 4.6 + 7.2i) = (2.5 – 4.6) + (3.8i + 7.2i) = –2.1 + 11i Distribute the negative sign. Combine the real parts. Combine the imaginary parts.

Operations with Complex Numbers Because i = , other powers of i can be calculated by considering the following pattern, i = i 5 = 1i = i i 2 = = –1 i 6 = –1 i 3 = = = –i i 7 = –i i 4 = i 2i 2 = (–1)(–1) = 1 i 8 = 1

Operations with Complex Numbers In most problems we will deal with, the most important power of i that you should know and use is i 2 = –1. When you multiply complex numbers, using this fact will help you to reduce the answers to complex form. Whenever you see i 2 in a calculation, you should replace it with a – 1 and continue to combine like terms and simplify.

Operations with Complex Numbers

Example 4 – Multiplying complex numbers Multiply the following complex numbers. a. 3(4 + 9i) b. 2i(7 – 3i) c. (2 + 5i)(4 + 8i) d. (3 + 2i)(3 – 2i)

Example 4 – Solution In all of these problems, we will use the distributive property and simplify where possible. a. 3(4 + 9i) = 12 + 27i b. 2i(7 – 3i) = 14i – 6i 2 = 14i – 6(–1) = 14i + 6 = 6 + 14i Distribute the 3. Distribute the 2i. Replace i 2 with –1. Simplify. Put in standard complex form a + bi.

Example 4 – Solution c. (2 + 5i)(4 + 8i) = 8 + 16i + 20i + 40i 2 cont’d c. (2 + 5i)(4 + 8i) = 8 + 16i + 20i + 40i 2 = 8 + 36i + 40i 2 = 8 + 36i + 40(–1) = 8 + 36i – 40 = –32 + 36i Use the distributive property (FOIL). Combine like terms. Replace i 2 with –1. Simplify.

Example 4 – Solution d. (3 + 2i)(3 – 2i) = 9 – 6i + 6i – 4i 2 cont’d d. (3 + 2i)(3 – 2i) = 9 – 6i + 6i – 4i 2 = 9 – 4i 2 = 9 – 4(–1) = 9 + 4 = 13 Use the distributive property (FOIL). Combine like terms. Replace i 2 with –1. Simplify.

Operations with Complex Numbers Part d of Example 4 is an example of multiplying two complex numbers that have a very special relationship. These two complex numbers are what are called complex conjugates of one another. When complex conjugates are multiplied together, note that the product is a real number. Therefore, it has no imaginary part remaining.

Operations with Complex Numbers

Example 6 – Multiplying by complex conjugates Multiply the following complex numbers by their conjugates. a. 2 + 9i b. 4 – 6i c. 3.4i

Example 6(a) – Solution (2 + 9i)(2 – 9i) = 4 – 18i + 18i – 81i 2 = 4 – 81(–1) = 4 + 81 = 85 Use the distributive property. Combine like terms. Replace i 2 with –1. Simplify.

Example 6(b) – Solution (4 – 6i)(4 + 6i) = 16 + 24i – 24i – 36i 2 cont’d (4 – 6i)(4 + 6i) = 16 + 24i – 24i – 36i 2 = 16 – 36i 2 = 16 – 36(–1) = 16 + 36 = 52 Use the distributive property. Combine like terms. Replace i 2 with –1. Simplify.

Example 6(c) – Solution 3.4i (–3.4i) = –11.56i 2 = –11.56(–1) = 11.56 cont’d 3.4i (–3.4i) = –11.56i 2 = –11.56(–1) = 11.56 Distribute. Replace i 2 with –1. Simplify.

Operations with Complex Numbers

Example 7 – Dividing complex numbers Divide the following. Give answers in the standard form for a complex number.

Example 7(a) – Solution Since the denominator does not have an imaginary part, we use the standard division algorithm. Reduce the fraction and put it into the standard form of a complex number. Simplify.

Example 7(b) – Solution cont’d Multiply the numerator and denominator by the conjugate of the denominator. Use the distributive property (FOIL).

Example 7(b) – Solution cont’d Write in the standard form of a complex number.

Example 7(c) – Solution cont’d Multiply the numerator and denominator by the conjugate of the denominator. Use the distributive property (FOIL).

Example 7(c) – Solution cont’d Write in the standard form of a complex number.

Example 7(d) – Solution cont’d Multiply the numerator and denominator by i. Because the denominator does not have a real part, multiply by i to rationalize the denominator. Write in the standard form of a complex number.

Solving Equations with Complex Solutions

Solving Equations with Complex Solutions Some equations will have complex solutions. The most common place in which we will see these types of solutions is in working with quadratics. The quadratic formula is a great tool to find both real and complex solutions to any quadratic equation. Now when a discriminant (b2 – 4ac) is a negative number, we can write our solutions using complex numbers instead of just saying that there are no real solutions. This results in a more complete answer to the equation.

Solving Equations with Complex Solutions Notice that if a complex number is a solution to a polynomial, the complex conjugate will also be a solution to that equation.

Example 8 – Solving equations with complex solutions Solve the following equations. Give answers in the standard form for a complex number. a. t2 + 2t + 5 = 0 b. x2 = –25 c. x2 + 4x = –30 d. x3 – 10x2 + 29x = 0

Example 8(a) – Solution Use the quadratic formula. The discriminant is –16, so the answer will be a complex number. Write in standard form.

Example 8(b) – Solution cont’d Use the square root property.

Example 8(c) – Solution cont’d Use the quadratic formula. The discriminant is –104, so the answer will be a complex number.

Example 8(c) – Solution cont’d Write in standard form.

Example 8(d) – Solution cont’d Factor the common term out. Set each factor equal to zero and continue to solve. Use the quadratic formula.

Example 8(d) – Solution cont’d The discriminant is –16, so the answer will be a complex number. Write in standard form.