What is #i^2#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Nam D. Apr 23, 2018 #-1# Explanation: The imaginary number, #i#, is defined as: #i=sqrt(-1)#. So, #i^2=(sqrt(-1))^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^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? How do you simplify #i^-33#? See all questions in Powers of Complex Numbers Impact of this question 73509 views around the world You can reuse this answer Creative Commons License