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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 $$ \blue{-28i + 28i} $$.

Video Tutorial on Dividing Complex Numbers


Practice Problems

Problem 1.1

Divide the complex numbers below:

$ \frac{ 3 + 5i}{ 2 + 6i} $

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

Find the following quotient

$ \frac{6-2i}{5 + 7i} $

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

Find the following quotient

$ \frac{3-2i}{3+2i} $

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...

Problem 1.3.1

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 ).

Exclamation Mark
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

Find the following quotient

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

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

Find the following quotient

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

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?

After looking at problems 1.5 and 1.6 , do you think that all complex quotients of the form

$ \frac{ \red a - \blue{ bi}}{\blue{ bi} - \red { a} } $

are equivalent to $$ -1$$? ( taken from our free downloadable worksheet )

The answer is yes

Any rational-expression in the form $$ \frac{y-x}{x-y} $$ is equivalent to $$-1$$.

(with the caveat that $$y-x \ne 0 $$).


Next to Complex Number Calc