How to multiply complex numbers. These binomials can be multiplied with FOIL

# Multiply Complex Numbers

How to multiply complex numbers, 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 $$-1$$.

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

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

### 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

 35 60i 14i + (-24) $$\bf{ 11+ 74i}$$

### Practice Problems I

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

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

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

 6 10i -12i + (20 ) 26 - 2i

### 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

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

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

 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

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

 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
Step 1

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

Use the shortcut to rewrite the left side

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

### Ultimate Math Solver (Free)

Free Algebra Solver ... type anything in there!