Multiply Complex Numbers

How to multiply complex numbers, complex conjugates

To multiply to complex numbers such as (4+5i )(3+2i) , you can treat each one as a binomial and apply the FOIL method to find the product.

FOIL
  • multiply the firsts 4 ×3 = 12
  • multiply the outers 4 × 2i = 8 i
  • multiply the inners 5i ×3 = 15i
  • multiply the lasts 5 i × 2 i = 10i 2= -10
Then add: 

12 + 4i + 15i -10 = 2 +23i

Video Tutorial on Multiplying Complex Numbers

Complex Conjugate

Complex conjugates are any pair of complex number binomials that look like the following pattern: (a + bi) & (a-bi)

Here are some specific examples. Note that the only difference between the two binomials is the sign.

  • (3 - 2i )( 3 +2i)
  • (5 + 12i )( 5 - 12i)
  • (7 - 33 i )( 7 +33i)

Multiplying complex conjugates causes the middle term to cancel as example 2 illutrates.

Example 1

Let's multiply the following 2 complex numbers (5 + 2i) (7 + 12i)

Step 1 Foil the binomials.
F : (5 + 2i ) (7 + 12i) 5 × 7 35
O : (5 + 2i ) (7 + 12i) 5 × 12i 60i
I : (5 + 2i ) (7 + 12i) 2i × 7 14i
L : (5 + 2i ) (7 + 12i) 2i × 12i 24i2 = -24 (remember i2 = -1)
Step 2

Simplify by adding the terms

35
60i
14i
+ (-24)
11+ 74i
Example 2

Let's multiply 2 complex conjugates (4 + 6i)(4 - 6i)

Step 1 Foil the binomials
F : (4 + 6i ) (4 - 6i) 4 × 4 16
O : (4 + 6i ) (4 - 6i) 4 × -6i -24i
I : (4 + 6i ) (4 - 6i) 6i × 4 24i
L : (4 + 6i ) (4 - 6i) 6i × 6i 36i2 = -36 (remember i2 = -1)
Step 2

Simplify by adding the terms

16
-24i
24i
+ (-36)
52 (notice how the imaginary terms are additive inverses or 'cancel' each other)

Practice Problems

Problem 1

Let's multiply the following complex numbers (5 + 4i) (6 + 4i)

Step 1

This problem is like example 1

F : (5 + 4i ) (6 + 4i) 5 × 6 30
O : (5 + 4i ) (6 + 4i) 5 × 4i 20i
I : (5 + 4i ) (6 + 4i) 4i × 6 24i
L : (5 + 4i ) (6 + 4i) 4i ×4i 16i2 = -16 (remember i2 = -1)
Step 2

Simplify by adding the terms

30
20i
24i
+ (-16)
14+ 44i
Problem 2

What is the product of the following complex numbers? (9 + 7i ) (6 + 8i)

Step 1

This problem is like example 1

F : (9 + 7i ) (6 + 8i) 9 × 6 54
O : (9 + 7i ) (6 + 8i) 9 × 8i 72i
I : (9 + 7i ) (6 + 8i) 7i × 6 42i
L : (9 + 7i ) (6 + 8i) 7i × 8i 56i2 = -56 (remember i2 = -1)
Step 2

Simplify by adding the terms

54
72i
42i
+ (-56)
-2 + 114i
Problem 3

Multiply the following complex numbers(5 + 4i ) (5 - 4i)

Step 1

This problem is like example 2 because the two binomials are complex conjugates.

F : (5 + 4i ) (5 - 4i) 5 × 5 25
O : (5 + 4i ) (5 - 4i) 5 × -4i -20i
I : (5 + 4i ) (5 - 4i) 4i × 5 20i
L : (5 + 4i ) (5 - 4i) 4i × -4i -16i2 = 16
Step 2

Simplify by adding the terms

25
-20i
20i
+ 16
41
Problem 4

Multiply the following complex numbers (6 + 2i ) (6 - 2i)

Step 1

This problem is like example 2 because the two binomials are complex conjugates.

F : (6 + 2i ) (6 - 2i) 6 × 6 36
O : (6 + 2i ) (6 - 2i) 6 × -2i -12i
I : (6 + 2i ) (6 - 2i) 2i × 6 12i
L : (6 + 2i ) (6 - 2i) 2i × -2i -4i2 = 4
Step 2

Simplify by adding the terms

36
-12i
12i
+ 6
40