How do you simplify #i^94#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Alan P. Feb 7, 2016 #i^94 =- 1# Explanation: #94 = 2xx47# #rarr i^94 = (i^2)^47 = (-1)^47# #(-1)^47 = (-1)^46xx(-1)# #color(white)("XXXX")=((-1)^2)^23)xx(-1)# #color(white)("XXXX")=(1^23)xx(-1)# #color(white)("XXXX")=1xx(-1)# #color(white)("XXXX")=-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 2387 views around the world You can reuse this answer Creative Commons License