How do you divide #(-2-9i)/(-2+7i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Ratnaker Mehta Aug 6, 2016 #-59/53+32/53i# Explanation: #(-2-9i)/(-2+7i)=(-2-9i)/(-2+7i) xx(-2-7i)/(-2-7i)# #={-2(-2-7i)-9i(-2-7i)}/{(-2)^2-(-7i)^2]# #=(4+14i+18i+63i^2}/{4-(49i^2}# #={4+32i+63(-1)}/{4-49(-1)}# #=(32i-59)/53# #=32/53i-59/53#. 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 1771 views around the world You can reuse this answer Creative Commons License