How do you simplify #i14#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer mason m Dec 9, 2015 #-1# Explanation: Rewrite #i^14# as #(i^4)^3xxi^2#. If #i=sqrt(-1),# then #i^2=-1#. From here #(i^2)^2=(-1)^2#, so #i^4=1#. If #i^4=1#, then we can say that: #i^14=(1)^3xxi^2# #=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 13863 views around the world You can reuse this answer Creative Commons License