How do you simplify #i^25#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Konstantinos Michailidis Nov 18, 2015 It is #i^25=(i^2)^12*i=(-1)^12*i=i# Assuning that #i# is the imaginary unit with #i^2=-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 36570 views around the world You can reuse this answer Creative Commons License