How do you simplify #i^333#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Lucio Falabella Jan 13, 2016 #i^333=i# Explanation: Using the exponent rules: #a^(n+m)=a^n*a^m# #a^(n*m)=(a^n)^m# #i^333=i*i^332=i*(i^2)^(332/2)=i*(i^2)^166# as #i^2=-1# #:.i*(i^2)^166=i*(-1)^166# as #(-1)^(2p)=1# #:.i*(-1)^166=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 4227 views around the world You can reuse this answer Creative Commons License