How do you divide #(-7+6i)/(10+5i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Ratnaker Mehta Aug 16, 2016 #=-8/25+(19i)/25# Explanation: #(-7+6i)/(10+5i)=(-7+6i)/(5(2+i)) xx (2-i)/(2-i)# #=((-7+6i)(2-i))/(5(2+i)(2-i))# #=(-14+7i+12i-6i^2)/(5(2^2-i^2))# #=(-14+19i-6(-1))/(5(4-(-1))# #=(-8+19i)/(5(5)# #=-8/25+(19i)/25# 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 1257 views around the world You can reuse this answer Creative Commons License