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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

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

Problem 1.1
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.

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

$$\red { [1]}$$ Remember $$i^2 = -1$$

Problem 1.2
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.

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

$$\red { [1]}$$ Remember $$i^2 = -1$$

Problem 1.3
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.

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

$$\red { [1]}$$ Remember $$i^2 = -1$$

More Like Problem 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?

Problem 1.4

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

Problem 1.5

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

Problem 1.4
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.

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

$$\red { [1]}$$ Remember $$i^2 = -1$$

Problem 1.5
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.

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

$$\red { [1]}$$ 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 $$y-x \ne 0$$).