How do you simplify #(-7+6i)/(9-4i)# and write the complex number in standard form? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Henry W. Oct 23, 2016 #(-87+26i)/85# Explanation: #(-7+6i)/(9-4i)*(9+4i)/(9+4i)->#multiplying by the conjugate of denominator #=((-7+6i)(9+4i))/(81-4i^2)=(-63-28i+54i-24)/(81+4)# #=(-87+26i)/85# 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 2567 views around the world You can reuse this answer Creative Commons License