How do you simplify #i^3#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Shwetank Mauria Jul 30, 2016 #i^3=-i# Explanation: As #i=sqrt(-1)# or #i^2=-1# Hence #i^3=ixxixxi=i^2xxi=-1xxi=-i# Answer link Related questions How do I multiply complex numbers? How do I multiply complex numbers in polar form? What is the formula for multiplying complex numbers in trigonometric form? How do I use the modulus and argument to square #(1+i)#? What is the geometric interpretation of multiplying two complex numbers? What is the product of #3+2i# and #1+7i#? How do I use DeMoivre's theorem to solve #z^3-1=0#? How do I find the product of two imaginary numbers? How do you simplify #(2+4i)(2-4i)#? How do you multiply #(-2-8i)(6+7i)#? See all questions in Multiplication of Complex Numbers Impact of this question 13060 views around the world You can reuse this answer Creative Commons License