How do you simplify #i^35#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Alan P. Nov 16, 2015 #i^35 = -i# Explanation: #i^2=-1# #i^4 = (i^2)^2 = (-1)^2 = +1# #i^35 = i^32*i^2*i# #= (i^4)^8*i^2*i# #=1^8*(-1)*i# #=-i# Answer link Related questions How do I use DeMoivre's theorem to find #(1+i)^5#? How do I use DeMoivre's theorem to find #(1-i)^10#? How do I use DeMoivre's theorem to find #(2+2i)^6#? What is #i^2#? What is #i^3#? What is #i^4#? How do I find the value of a given power of #i#? How do I find the #n#th power of a complex number? How do I find the negative power of a complex number? Write the complex number #i^17# in standard form? See all questions in Powers of Complex Numbers Impact of this question 44562 views around the world You can reuse this answer Creative Commons License