How do you simplify #i^6#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Konstantinos Michailidis Dec 27, 2016 We know that #i^2=-1# hence #i^6=(i^2)^3=i^2*i^2*i^2=(-1)*(-1)*(-1)=-1# 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 7492 views around the world You can reuse this answer Creative Commons License