How do you simplify #(i^2 + 2i^3) / i#? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 1 Answer Shwetank Mauria Mar 9, 2016 #(i^2+2i^3)/i=-2+i# Explanation: As (i^2=-1#, #(i^2+2i^3)/i=i+2i^2=i+2(-1)=i-2=-2+i# 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 1614 views around the world You can reuse this answer Creative Commons License