How do you simplify # (3i)/(1+i) + 2/(2+3i) #? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Tony B Nov 5, 2015 #(-7 +12i)/( -1 + 5i)# Explanation: Use a common denominator: # ( 3i(2+3i) + 2(1+i))/((1+i)(2+3i))# #(6i +9(i)^2 +2 +2i)/(2 + 3i +2i + 3(i)^2# But #(i)^2 = -1# giving: #(-7 +8i)/( -1 + 5i)# 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 1232 views around the world You can reuse this answer Creative Commons License