How do you simplify #i^7#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Alan P. Feb 7, 2016 #i^7= -i# Explanation: #i^7 = (i^2)^3xxi# #color(white)("XXX")=(-1)^3xxi# #color(white)("XXX")=(-1)^2xx(-1)xxi# #color(white)("XXX")=1xx(-1)xxi# #color(white)("XXX")=-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 12418 views around the world You can reuse this answer Creative Commons License