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

How to Divide Complex Numbers

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

Video Tutorial on Dividing Complex Numbers


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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}} $

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

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} $

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

Problem 1.3

Find the quotient of

$ \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 } \\ $

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

Another question like 1.3 ...
Problem 1.4
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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 }} $

Problem 1.5

Find the quotient of

$ \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} $

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

Problem 1.6

Find the quotient of

$ \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} $

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

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 $$y-x \ne 0 $$)


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