How do I divide complex numbers in standard form?

2 Answers
May 23, 2015

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#

May 23, 2015

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)#.