# How do I divide complex numbers in standard form?

May 23, 2015

Assuming you have two complex numbers $a + b i$ and $c + \mathrm{di}$
you can start by treating the constant $i$ as if it were a variable $x$
and you were asked to evaluate
$\left(a + b x\right) \div \left(c + \mathrm{dx}\right)$
In most cases you will end up with a remainder which needs have it's denominator converted into a non-complex form.

This is easier to see if we work with an example:
(Warning: I've made up these numbers on the fly, so the answer is likely to be ugly)

Suppose you were asked to divide:
$\left(6 + 7 i\right) \div \left(2 + 3 i\right)$

Either through observation or synthetic division you could obtain
$\left(6 + 7 i\right) \div \left(2 + 3 i\right)$
$= 3 + \text{Remainder of } \left(- 2 i\right)$
or
$= 3 + \frac{- 2 i}{2 + 3 i}$

We wish to clear the complex component of the denominator, so
$= 3 + \frac{- 2 i}{2 + 3 i} \times \frac{2 - 3 i}{2 - 3 i}$

$= 3 + \frac{- 4 i - 6}{4 + 9}$

$= 3 - \frac{6}{13} - \frac{4 i}{13}$

$= \frac{33}{13} - \frac{4}{13} i$

May 23, 2015

In this way, remembering that ${i}^{2} = - 1$ and given $z = a + i b$ and $w = c + i d$, then:

$\frac{z}{w} = \frac{a + i b}{c + i d} = \frac{a + i b}{c + i d} \cdot \frac{c - i d}{c - i d} = \frac{\left(a + i b\right) \left(c - i d\right)}{{c}^{2} - {\left(i d\right)}^{2}} =$

$= \frac{a c - i a d + i b c - {i}^{2} b d}{{c}^{2} - {i}^{2} {d}^{2}} = \frac{a c - i a d + i b c + b d}{{c}^{2} + {d}^{2}} =$

$= \frac{a c + b d}{{c}^{2} + {d}^{2}} + i \frac{b c - a d}{{c}^{2} + {d}^{2}}$.