How do you simplify #(4-6i)/i#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Ratnaker Mehta Sep 8, 2016 #(4-6i)/i=(2(2-3i))/i=2(2/i-3i/i)=2(2/ixxi/i-3)# #=2((2i)/i^2-3)=2((2i)/-1-3)=-2(2i+3), or, -2(3+2i)#. 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 2369 views around the world You can reuse this answer Creative Commons License