How do you simplify #i^57#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer José F. Mar 12, 2016 #i# Explanation: #color(red)(i^4=1)# #i^57=(i^4)^(14)*i=1^14*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 15492 views around the world You can reuse this answer Creative Commons License