How do you simplify #i^266#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer mason m · Alan P. Jan 20, 2016 #-1# Explanation: The powers of #i# go as follow: #i=sqrt(-1)# #i^2=-1# #i^4=(i^2)^2=1# Thus, we can rewrite #i^266# in terms of these: #i^266=(i^4)^66(i^2)=(1)^66(-1)=-1# 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 1599 views around the world You can reuse this answer Creative Commons License