How do you simplify #(3-2i)/(-4-i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Lithia Feb 17, 2017 #(-10+11i)/17# Explanation: #(3-2i)/(-4-i) * (-4+i)/(-4+i)# multiply by its conjugate remember that #color(orange)(i^2=-1)# #(-12+3i+8i-2i^2)/(16-4i+4i-i^2)=(-12+11i-2(-1))/(16-(-1))=(-10+11i)/17# 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 10879 views around the world You can reuse this answer Creative Commons License