Multiply Complex Numbers
How to multiply complex numbers, complex conjugates
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 × 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) |
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| Step 2) Simplify by adding the terms |
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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) |
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| Step 2) Simplify by adding the terms |
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52 (notice how the imaginary terms are additive inverses or 'cancel' each other) |
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Practice Problems
Problem 1)Let's multiply the following complex numbers
(5 + 4i ) (6 + 4i)
This problem is like example 1.
| Step 1) Foil the binomials. |
| 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) |
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| Step 2) Simplify by adding the terms |
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Problem 2) What is the product of the following complex numbers?
(9 + 7i ) (6 + 8i)
This problem is like example 1.
| Step 1) Foil the binomials. |
| 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) |
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| Step 2) Simplify by adding the terms |
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Problem 3) multiply the following complex numbers
(5 + 4i ) (5 - 4i)
This problem is like example 2 because the two binomials are complex conjugates.
| Step 1) Foil the binomials. |
| 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 |
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| Step 2) Simplify by adding the terms |
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Problem 4) multiply the following complex numbers
(6 + 2i ) (6 - 2i)
This problem is like example 2 because the two binomials are complex conjugates.
| Step 1) Foil the binomials. |
| 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 |
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| Step 2) Simplify by adding the terms |
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