How do you simplify #i^37#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Konstantinos Michailidis Feb 19, 2016 It is #i# Explanation: Remember that #i^2=-1# hence #i^37=(i^36)*i=(i^2)^18*i=(-1)^18*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 29587 views around the world You can reuse this answer Creative Commons License