To multiply two complex numbers such as $$\ (4+5i )\cdot (3+2i) $$, 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 $$ -1 $$.
Add up each term!
| $$ 6 $$ |
| $$ 18i $$ |
| $$ 21i $$ |
| $$ -63$$ |
| $$ \bf{39i - 63} $$ |
Let's multiply the following 2 complex numbers $$ \bf{ (5 + 2i) (7 + 12i)} $$
Step 1 Foil the binomials.$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 5 + 2i)( \red 7 +12i ) & 5 \cdot 7 & 35 \\\hline \bf{O} & ( \red 5 + 2i)( 7 +\red{12i } ) & 5 \cdot 12i & 60i \\\hline \bf{I} & ( 5 + \red{ 2i})( \red 7 + 12i ) & 2i \cdot 7 & 14i \\\hline \bf{L} & ( 5 + \red{ 2i})( 7 +\red{12i } ) & 5i \cdot 12i & 24i^2 = -24 \\\hline \end{array} $$
Remember $$ i^2 = -1 $$, so $$ 24i^2 = 24 \cdot -1 = -24 $$
Step 2Simplify by adding the terms
| $$ 35 $$ |
| $$ 60i $$ |
| $$ 14i $$ |
| $$ + (-24) $$ |
| $$ \bf{ 11+ 74i} $$ |
This problem is like example 1
$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 5 + 4i)( \red 6 + 4i ) & 5 \cdot 6 & 30 \\\hline \bf{O} & ( \red 5 + 4i)( 6 +\red{4i } ) & 5 \cdot 4i & 20i \\\hline \bf{I} & ( 5 + \red{ 4i})( \red 6 + 4i ) & 4i \cdot 6 & 24i \\\hline \bf{L} & ( 5 + \red{ 4i})( 6 +\red{4i } ) & 4i \cdot 4i & -16_{\blue{ [1]}} \\\hline \end{array} $$
$$ [\blue 1] $$ Remember $$ i^2 = -1 $$, so $$ 4i \cdot 4i = 16i^2 = 16 \cdot -1 = -16 $$
Simplify by adding the terms
| $$30$$ |
| $$20i$$ |
| $$24i$$ |
| $$+ (-16)$$ |
| $$\bf{14+ 44i}$$ |
This problem is like example 1
$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 9 + 7i)( \red 6 + 8i ) & 9 \cdot 6 & 54 \\\hline \bf{O} & ( \red 9 + 7i)( 6 +\red{8i } ) & 9 \cdot 8i & 72i \\\hline \bf{I} & ( 9 + \red{ 7i})( \red 6 + 8i ) & 7i \cdot 6 & 42i \\\hline \bf{L} & ( 9 + \red{ 7i})( 6 +\red{ 8i } ) & 7i \cdot 8i & 56i^2 = -56 \\\hline \end{array} $$
Simplify by adding the terms
| $$54$$ |
| $$72i$$ |
| $$42i$$ |
| $$+ (-56)$$ |
| $$\bf{-2 + 114i}$$ |
This problem is like example 1
$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 2 - 4i)( \red 3 + 5i ) & 2 \cdot 3 & 6 \\\hline \bf{O} & ( \red 2 - 4i)( 3 +\red{5i } ) & 2 \cdot 5i & 10i \\\hline \bf{I} & ( 2 - \red{ 4i})( \red 3 + 5i ) & -4i \cdot 3 & -12i \\\hline \bf{L} & ( 2 - \red{ 4i})( 3 +\red{ 5i } ) & -4i \cdot 5i & -20i^2 = 20 \\\hline \end{array} $$
Remember $$ i^2 = -1 $$, so $$ -20i^2 = -20 \cdot -1 = 20 $$
Simplify by adding the terms
| $$6$$ |
| $$10i$$ |
| $$12i$$ |
| $$+ (20)$$ |
| $$]bf{26 - 2i}$$ |
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.
Let's multiply 2 complex conjugates $$ (4 + 6i)(4 - 6i) $$
Step 1 Foil the binomials.$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 4 + 6i)( \red 4 - 6i ) & 4 \cdot 4 & 16 \\\hline \bf{O} & ( \red 4 + 6i)( 4 -\red{6i } ) & 4 \cdot -6i & -24i \\\hline \bf{I} & ( 4 + \red{ 6i})( \red 4 - 6i ) & 6i \cdot 4 & 24i \\\hline \bf{L} & ( 4 + \red{ 6i})( 4 - \red{6i } ) & 6i \cdot -6i & 36_{\blue{ [1]}} \\\hline \end{array} $$
$$[\blue{1}]$$ Remember $$ i^2 = -1 $$, so $$ -36i^2 = -36 \cdot -1 = 36 $$
Step 2Simplify by adding the terms
| $$ 16 $$ |
| $$ \red{-24i} $$ |
| $$ \red{24i} $$ |
| $$ + 36 $$ |
| $$ \bf{52} $$ |
(notice how the imaginary terms are additive inverses or 'cancel' each other)
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} } $
This problem is like example 2 because the two binomials are complex conjugates.
$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 5 + 4i)( \red 5 - 4i ) & 5 \cdot 5 & 25 \\\hline \bf{O} & ( \red 5 + 4i)( 5 -\red{4i } ) & 5 \cdot \red- 4i & \red-20i \\\hline \bf{I} & ( 5 + \red{ 4i})( \red 5 - 4i ) & 4i \cdot 5 & 20i \\\hline \bf{L} & ( 5 + \red{ 4i})( 5 - \red{ 4i } ) & 4i \cdot \red-4i & -16i^2 = 16 \\\hline \end{array} $$
Remember $$ i^2 = -1 $$, so $$ -16i^2 = -16 \cdot -1 = 16 $$
Simplify by adding the terms
| $$25$$ |
| $$\red{-20i}$$ |
| $$\red{20i}$$ |
| $$+ 16$$ |
| $$\bf{41}$$ |
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
This problem is like example 2 because the two binomials are complex conjugates.
$$ \begin{array}{| c|c | c | } \bf{F} & ( \red 6 + 2i)( \red 6 - 2i ) & 6 \cdot 6 & 36 \\\hline \bf{O} & ( \red 6 + 2i)( 6 -\red{2i } ) & 6 \cdot \red- 2i & \red-12i \\\hline \bf{I} & ( 6 + \red{ 2i})( \red 6 - 2i ) & 2i \cdot 6 & 12i \\\hline \bf{L} & ( 6 + \red{ 2i})( 6 - \red{ 2i } ) & 2i \cdot \red-2i & -4i^2 = 4 \\\hline \end{array} $$
Simplify by adding the terms
| $$36$$ |
| $$\red{-12i}$$ |
| $$\red{12i}$$ |
| $$+ 4$$ |
| $$\bf{40}$$ |
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
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 $
