To multiply to complex numbers such as (4+5*i* )(3+2*i*) , 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 × 2
*i*= 8*i* - multiply the inners 5
*i*×3 = 15*i* - multiply the lasts 5
*i*× 2*i*= 10*i*^{ 2}= -10

**Then add:**

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

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

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

##### 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 | 24i^{2} = -24 (remember i^{2} = -1) |

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 binomialsF : (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 | 36i^{2} = -36 (remember i^{2} = -1) |

Simplify by adding the terms

16 |

-24i |

24i |

+ (-36) |

52 (notice how the imaginary terms are additive inverses or 'cancel' each other) |

**Practice** Problems

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 | 16i^{2} = -16 (remember i^{2} = -1) |

Simplify by adding the terms

30 |

20i |

24i |

+ (-16) |

14+ 44i |

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 | 56i^{2} = -56 (remember i^{2} = -1) |

Simplify by adding the terms

54 |

72i |

42i |

+ (-56) |

-2 + 114i |

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 | -16i^{2} = 16 |

Simplify by adding the terms

25 |

-20i |

20i |

+ 16 |

41 |

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 | -4i^{2} = 4 |

Simplify by adding the terms

36 |

-12i |

12i |

+ 6 |

40 |