﻿ Divide Complex Numbers: How to divide complex numbers using the complex conjugate to rationalize the denominator.

# How to Divide Complex Numbers

#### It's All about complex conjugates and multiplication

To divide complex numbers. First, find the complex conjugate of the denominator, multiply the numerator and denominator by that conjugate and simplify.
##### Example 1

Let's divide the following 2 complex numbers

$\frac{ 5 + 2i}{ 7 + 4i}$

Step 1

Determine the conjugate of the denominator

The conjugate of $$(7 + 4i)$$ is $$(7 \red - 4i)$$

Step 2

Multiply the numerator and denominator by the conjugate.

$\big( \frac{ 5 + 2i}{ 7 + 4i} \big) \big( \frac{ 7 \red - 4i}{7 \red - 4i} \big)$

Step 3

Simplify

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

$\big( \frac{ 5 + 2i}{ 7 + 4i} \big) \big( \frac{ 7 \red - 4i}{7 \red - 4i} \big) \\ \boxed{ \frac{ 35 + 14i -20i - 8\red{i^2 } }{ 49 \blue{-28i + 28i}-16 \red{i^2 }} } \\ \frac{ 35 + 14i -20i \red - 8 }{ 49 \blue{-28i + 28i} \red -16 } \\ \frac{ 43 -6i }{ 65 }$

Note: The reason that we use the complex conjugate of the denominator is so that the $$i$$ term in the denominator "cancels", which is what happens above with the i terms highlighted in blue $$\blue{-28i + 28i}$$.

### Practice Problems

Step 1

Determine the conjugate of the denominator

The conjugate of $$2 + 6i$$ is $$(2 \red - 6i)$$

Step 2

Multiply the numerator and denominator by the conjugate.

$\big( \frac{ 3 + 5i}{ 2 + 6i} \big) \big( \frac { 2 \red - 6i}{ 2 \red - 6i} \big)$

Step 3

Simplify.

$\big( \frac{ 3 + 5i}{ 2 + 6i} \big) \big( \frac { 2 \red - 6i}{ 2 \red - 6i} \big) \\ \frac{ 6 -18i +10i -30 \red{i^2} }{ 4 \blue{ -12i+12i} -36\red{i^2}} \text{ } _{ \small{ \red { [1] }}} \\ \frac{ 6 -8i \red + 30 }{ 4 \red + 36}= \frac{ 36 -8i }{ 40 } \\ \boxed{ \frac{9 -2i}{10}}$

$$\red { [1]} (\text{ remember } i^2 = -1)$$

Step 1

Determine the conjugate of the denominator

The conjugate of $$5 + 7i$$ is $$5 \red - 7i$$

Step 2

Multiply the numerator and denominator by the conjugate.

$\big( \frac{6-2i}{5 + 7i} \big) \big( \frac{5 \red- 7i}{5 \red- 7i} \big)$

Step 3

Simplify

$\big( \frac{6-2i}{5 + 7i} \big) \big( \frac{5 \red- 7i}{5 \red- 7i} \big) \\ \frac{ 30 -42i - 10i + 14\red{i^2}}{25 \blue{-35i +35i} -49\red{i^2} } \text{ } _{ \small{ \red { [1] }}} \\ \frac{ 30 -52i \red - 14}{25 \red + 49 } = \frac{ 16 =52i}{ 74}$

$$\red { [1]} (\text{ remember } i^2 = -1)$$

Step 1

Determine the conjugate of the denominator

The conjugate of $$3 + 2i$$ is $$(3 \red -2i)$$

Step 2

Multiply the numerator and denominator by the conjugate

$\big( \frac{ 3 -2i}{ 3 + 2i} \big) \big( \frac { 3 \red - 2i}{ 3 \red - 2i} \big)$

Step 3

Simplify

$\big( \frac{ 3 -2i}{ 3 + 2i} \big) \big( \frac { 3 \red - 2i}{ 3 \red - 2i} \big) \\ \frac{ 9 \blue{ -6i -6i }+ 4 \red{i^2 } }{ 9 \blue{ -6i +6i } - 4 \red{i^2 }} \text{ } _{ \small{ \red { [1] }}} \\ \frac{ 9 \blue{ -12i } -4 }{ 9 + 4 } \\ \frac{ 5 -12i }{ 13 } \\$

$$\red { [1]} (\text{ remember } i^2 = -1)$$

Another question like 1.3 ...

#### Make a Prediction

Look carefully at the problems 1.5 and 1.6 below.

Make a Prediction: Do you think that there will be anything special or interesting about either of the following quotients? Scroll down the page to see the answer. ( from our free downloadable worksheet )

$\frac{ \red 3 - \blue{ 2i}}{\blue{ 2i} - \red { 3} }$

$\frac{\red 4 - \blue{ 5i}}{\blue{ 5i } - \red{ 4 }}$

Step 1

Determine the conjugate of the denominator

The conjugate of $$2i -3$$ is $$(2i \red + 3 )$$

Step 2

Multiply the numerator and denominator by the conjugate

$\big( \frac{ 3 -2i}{ 2i -3 } \big) \big( \frac { 2i \red + 3 }{ 2i \red + 3 } \big)$

Step 3

Simplify

$\big( \frac{ 3 -2i}{ 2i -3 } \big) \big( \frac { 2i \red + 3 }{ 2i \red + 3 } \big) \\ \frac{ \blue{6i } + 9 - 4 \red{i^2 } \blue{ -6i } }{ 4 \red{i^2 } + \blue{6i } - \blue{6i } - 9 } \text{ } _{ \small{ \red { [1] }}} \\ \frac{ 9 + 4 }{ -4 - 9 } \\ \boxed{-1}$

$$\red { [1]} (\text{ remember } i^2 = -1)$$

Step 1

Determine the conjugate of the denominator

The conjugate of $$5i - 4$$ is $$(5i \red + 4 )$$

Step 2

Multiply the numerator and denominator by the conjugate

$\big( \frac{ 4 -5i}{ 5i -4 } \big) \big( \frac { 5i \red + 4 }{ 5i \red + 4 } \big)$

Step 3

Simplify

$\big( \frac{ 4 -5i}{ 5i -4 } \big) \big( \frac { 5i \red + 4 }{ 5i \red + 4 } \big) \\ \frac{\blue{20i} + 16 -25\red{i^2} -\blue{20i}} { 25\red{i^2} + \blue{20i} - \blue{20i} -16} \text{ } _{ \small{ \red { [1] }}} \\ \frac{ 16 + 25 }{ -25 - 16 } \\ \frac{ 41 }{ -41 } \\ \boxed{-1}$

$$\red { [1]} (\text{ remember } i^2 = -1)$$

# How did we get the same result?

Any rational-expression in the form $$\frac{y-x}{x-y}$$is equivalent to $$-1$$
(with the caveat that $$y-x \ne 0$$)