Question

How do I divide complex numbers in standard form?

Answer 1

Assuming you have two complex numbers `a+bi` and `c+di`
you can start by treating the constant `i` as if it were a variable `x`
and you were asked to evaluate
`(a+bx) div (c+dx)`
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:
`(6+7i) div (2+3i)`

Either through observation or synthetic division you could obtain
`(6+7i) div (2+3i)`
`= 3 + "Remainder of "(-2i)`
or
`= 3+ (-2i)/(2+3i)`

We wish to clear the complex component of the denominator, so
`=3+(-2i)/(2+3i) xx(2-3i)/(2-3i)`

`= 3+ (-4i -6)/(4+9)`

`=3-6/13 -(4i)/13`

`=33/13-4/13i`

Answer 2

In this way, remembering that `i^2=-1` and given `z=a+ib` and `w=c+id`, then:

`z/w=(a+ib)/(c+id)=(a+ib)/(c+id)*(c-id)/(c-id)=((a+ib)(c-id))/(c^2-(id)^2)=`

`=(ac-iad+ibc-i^2bd)/(c^2-i^2d^2)=(ac-iad+ibc+bd)/(c^2+d^2)=`

`=(ac+bd)/(c^2+d^2)+i(bc-ad)/(c^2+d^2)`.