How do you divide #3/(-i)#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Rory K. Sep 15, 2016 3i Explanation: #3/-i# x #(-i)/-i# = #(-3i)/(-i)^2# #(-3i)/i^2# = 3i ....since (#[sqrt(-1)]^2# = -1 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 1498 views around the world You can reuse this answer Creative Commons License