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 1Determine the conjugate of the denominator
The conjugate of $$ (7 + 4i)$$ is $$ (7 \red - 4i)$$.
Step 2Multiply the numerator and denominator by the conjugate.
$ \big( \frac{ 5 + 2i}{ 7 + 4i} \big) \big( \frac{ 7 \red - 4i}{7 \red - 4i} \big) $
Step 3Simplify
$ \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{ 35 + 14i -20i \red - 8 }{ 49 \blue{-28i + 28i} +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
Step 1
Step 2
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}} $
Problem 1.2
Step 1
Step 2
Step 3
Simplify.
Problem 1.3
Step 1
Step 2
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 } \\ $
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 ).
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
Step 2
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} $
Problem 1.5
Step 1
Step 2
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} $
How did we get the same result?
The answer is yes
Any rational-expression in the form $$ \frac{y-x}{x-y} $$ is equivalent to $$-1$$.
(with the caveat that $$y-x \ne 0 $$).
