How do you divide #(-2+3i)/(3-2i)#?

1 Answer
Oct 28, 2016

Answer:

The answer is #-12/13+(5i)/13#

Explanation:

To simplify, multiply numerator and denominatorby the conjugate of the denominator
if #z=a+ib# the the conjugate is #barz=a-ib#
so the conjugate of the denominator is #3+2i#
and #i^2=-1#
Then the expression is #(-2+3i)/(3-21)=((-2+3i)(3+2i))/((3-2i)(3+2i))#

#=(-6-4i+9i+6i^2)/(9-4i^2)=(-12+5i)/(13)=-12/13+(5i)/13#