How do you simplify #( 3-2i)/(3+2i )#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer sente May 18, 2016 #(3-2i)/(3+2i)=5/13-12/13i# Explanation: Given a complex number #a+bi#, the complex conjugate of that number is #a-bi#. A useful property is that the product of any number with its complex conjugate is a real number. We will use that to eliminate the complex number from the denominator. #(3-2i)/(3+2i) = ((3-2i)(3-2i))/((3+2i)(3-2i))# #=(5-12i)/13# #=5/13-12/13i# Answer link Related questions How do I graphically divide complex numbers? How do I divide complex numbers in standard form? How do I find the quotient of two complex numbers in polar form? How do I find the quotient #(-5+i)/(-7+i)#? How do I find the quotient of two complex numbers in standard form? What is the complex conjugate of a complex number? How do I find the complex conjugate of #12/(5i)#? How do I rationalize the denominator of a complex quotient? How do I divide #6(cos^circ 60+i\ sin60^circ)# by #3(cos^circ 90+i\ sin90^circ)#? How do you write #(-2i) / (4-2i)# in the "a+bi" form? See all questions in Division of Complex Numbers Impact of this question 6037 views around the world You can reuse this answer Creative Commons License