How do you simplify #i^13#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer sjc Oct 21, 2016 #i# Explanation: #i=sqrt-1# #i^2=-1# #i^3=i# #i^4=1# multiplying bi #i# repeats the pattern. So any #i^(4n)=i# Since #13-=1(mod4)# #i^(13)=i^1=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 22059 views around the world You can reuse this answer Creative Commons License