foil complex binomials

Multiply Complex Numbers

How to multiply complex numbers, complex conjugates

This page: Lesson | Practice | Complex Conjugates

Just like multiplying binomials

To multiply two complex numbers such as $$\boxed{ (4+5i )\cdot (3+2i) }$$ , you can treat each one as a binomial and apply the foil method to find the product.

FOIL stands for first , outer, inner , and last pairs . You are supposed to multiply these pairs as shown below!

Firsts : $$ 2 \cdot 3= 6 $$
Outers : $$ 2 \cdot 9i =18i $$
Inners : $$ 7i \cdot 3 =21i $$
Lasts : $$ 7i \cdot 9i =$$ $$63i^2$$$$ = 63\cdot -1 = -63 $$

Note: the $$ i^2 $$ simplifies to $$ -1 $$.
FOIL explained binomials

So, now that we've multiplied, what is next?

Add up each term!

$$ 6 $$
$$ 18i $$
$$ 21i $$
$$ -63$$
$$ 39i - 63 $$

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Video Tutorial on Multiplying Complex Numbers

Example 1

Let's multiply the following 2 complex numbers $$ \bf{ (5 + 2i) (7 + 12i)} $$

Step 1 Foil the binomials.

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 5 + 2i)( \red 7 +12i ) & 5 \cdot 7 & 35 \\\hline \bf{O} & ( \red 5 + 2i)( 7 +\red{12i } ) & 5 \cdot 12i & 60i \\\hline \bf{I} & ( 5 + \red{ 2i})( \red 7 + 12i ) & 2i \cdot 7 & 14i \\\hline \bf{L} & ( 5 + \red{ 2i})( 7 +\red{12i } ) & 5i \cdot 12i & 24i^2 = -24 \\\hline \end{array} $$

$$ (\text{ remember } i^2 = -1 \text{ so } 24i^2 = 24 \cdot -1 = -24 ) $$

Step 2

Simplify by adding the terms

35
60i
14i
+ (-24)
$$ \bf{ 11+ 74i} $$
This page: Lesson | Practice | Complex Conjugates

Practice Problems I

Problem 1.1

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

Step 1

This problem is like example 1

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 5 + 4i)( \red 6 + 4i ) & 5 \cdot 6 & 30 \\\hline \bf{O} & ( \red 5 + 4i)( 6 +\red{4i } ) & 5 \cdot 4i & 20i \\\hline \bf{I} & ( 5 + \red{ 4i})( \red 6 + 4i ) & 4i \cdot 6 & 24i \\\hline \bf{L} & ( 5 + \red{ 4i})( 6 +\red{4i } ) & 4i \cdot 4i & -16_{\blue{ [1]}} \\\hline \end{array} $$

$$ [\blue 1] (\text{ remember } i^2 = -1 \text{ so } 4i \cdot 4i = 16i^2 = 16 \cdot -1 = -16 ) $$

Step 2

Simplify by adding the terms

30
20i
24i
+ (-16)
14+ 44i
Problem 1.2

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

Step 1

This problem is like example 1

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 9 + 7i)( \red 6 + 8i ) & 9 \cdot 6 & 54 \\\hline \bf{O} & ( \red 9 + 7i)( 6 +\red{8i } ) & 9 \cdot 8i & 72i \\\hline \bf{I} & ( 9 + \red{ 7i})( \red 6 + 8i ) & 7i \cdot 6 & 42i \\\hline \bf{L} & ( 9 + \red{ 7i})( 6 +\red{ 8i } ) & 7i \cdot 8i & 56i^2 = -56 \\\hline \end{array} $$

$$ (\text{ remember } i^2 = -1 \text{ so } 56i^2 = 56 \cdot -1 = -56 ) $$

Step 2

Simplify by adding the terms

54
72i
42i
+ (-56)
-2 + 114i
Problem 1.3

What is the product of the following complex numbers? $$ (2 - 4i ) (3 + 5i) $$ taken from our free downloadable worksheet

Step 1

This problem is like example 1

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 2 - 4i)( \red 3 + 5i ) & 2 \cdot 3 & 6 \\\hline \bf{O} & ( \red 2 - 4i)( 3 +\red{5i } ) & 2 \cdot 5i & 10i \\\hline \bf{I} & ( 2 - \red{ 4i})( \red 3 + 5i ) & -4i \cdot 3 & -12i \\\hline \bf{L} & ( 2 - \red{ 4i})( 3 +\red{ 5i } ) & -4i \cdot 5i & -20i^2 = 20 \\\hline \end{array} $$

$$ (\text{ remember } i^2 = -1 \text{ so } -20i^2 = -20 \cdot -1 = 20 ) $$

Step 2

Simplify by adding the terms

6
10i
-12i
+ (20 )
26 - 2i
foil complex binomials This page: Lesson | Practice | Complex Conjugates

Complex Conjugate

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

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

$ \text{Complex Conjugate Examples} $

$ \\(3 \red + 2i)(3 \red - 2i) \\(5 \red + 12i)(5 \red - 12i) \\(7 \red + 33i)(5 \red - 33i) \\(99 \red + i)(99 \red - i) $

Multiplying complex conjugates causes the middle term ( the $$i$$ term) to cancel as example 2 below illustrates.

Example 2 (Complex Conjugate)

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

Step 1 Foil the binomials

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 4 + 6i)( \red 4 - 6i ) & 4 \cdot 4 & 16 \\\hline \bf{O} & ( \red 4 + 6i)( 4 -\red{6i } ) & 4 \cdot -6i & -24i \\\hline \bf{I} & ( 4 + \red{ 6i})( \red 4 - 6i ) & 6i \cdot 4 & 24i \\\hline \bf{L} & ( 4 + \red{ 6i})( 4 - \red{6i } ) & 6i \cdot -6i & 36_{\blue{ [1]}} \\\hline \end{array} $$

$$ [\blue 1] (\text{ remember } i^2 = -1 \text{ so } -36i^2 = -36 \cdot -1 = 36 ) $$

Step 2

Simplify by adding the terms

16
-24i
24i
+36
52

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

Shortcut for Multiplying Complex Conjugates

There is a shortcut that you can use to quickly multiply complex conjugates .

As you can see from the last example

$ (4+6i)(4-6i) = 16 + 36 \\ (\red 4+ \blue 6i)(\red 4- \blue 6i) = \red{4^2} + \red{6^2} $


shortcut: $ (\red a + \blue b i)( \red a - \blue b i)= \boxed{ \red {a^2} + \blue {b^2} } $

Problem 2.1

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

Step 1

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

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 5 + 4i)( \red 5 - 4i ) & 5 \cdot 5 & 25 \\\hline \bf{O} & ( \red 5 + 4i)( 5 -\red{4i } ) & 5 \cdot \red- 4i & \red-20i \\\hline \bf{I} & ( 5 + \red{ 4i})( \red 5 - 4i ) & 4i \cdot 5 & 20i \\\hline \bf{L} & ( 5 + \red{ 4i})( 5 - \red{ 4i } ) & 4i \cdot \red-4i & -16i^2 = 16 \\\hline \end{array} $$

$$ (\text{ remember } i^2 = -1 \text{ so } 16i^2 = 16 \cdot -1 = 16 ) $$

Step 2

Simplify by adding the terms

25
-20i
20i
+ 16
41

shortcut: There is a shortcut that applies to complex conjugates of the form $$ (\red a + \blue b i)( \red a - \blue b i)$$ like this question. The solution is always $$ \red {a^2} + \blue {b^2} $$. In this case: $$ \red {5^2} + \blue {4^2} $$ = 41


Problem 2.2

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

Step 1

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

$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 6 + 2i)( \red 6 - 2i ) & 6 \cdot 6 & 36 \\\hline \bf{O} & ( \red 6 + 2i)( 6 -\red{2i } ) & 6 \cdot \red- 2i & \red-12i \\\hline \bf{I} & ( 6 + \red{ 2i})( \red 6 - 2i ) & 2i \cdot 6 & 12i \\\hline \bf{L} & ( 6 + \red{ 2i})( 6 - \red{ 2i } ) & 2i \cdot \red-2i & -4i^2 = 4 \\\hline \end{array} $$

$$ (\text{ remember } i^2 = -1 \text{ so } 16i^2 = 16 \cdot -1 = 16 ) $$

Step 2

Simplify by adding the terms

36
-12i
12i
+ 6
40
shortcut: There is a shortcut that applies to complex conjugates of the form $$ (\red a + \blue b i)( \red a - \blue b i)$$ like this question. The solution is always $$ \red {a^2} + \blue {b^2} $$. In this case: $$ \red {6^2} + \blue {2^2} $$ = 40
Problem 2.3

Find a, if (1 + ai ) (1 - ai) = 2 , taken from our free downloadable worksheet

Step 1

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

Use the shortcut to rewrite the left side

$ (\red 1 + \blue ai ) (\red 1 - \blue ai) = 2 \\ \red {1^2} + \blue {a^2} = 2 \\ 1 + a^2 = 2 \\ a^2 = 1 \\ a= \sqrt 1 \\ a= 1 $


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