How do you simplify #i^-11#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer sente Dec 19, 2015 Simplify the expression by using the fact that #i^4 = 1# to find #i^(-11) = i# Explanation: Using that #i^2 = -1# and thus #i^4 = (i^2)^2 = 1# we have #i^-11 = 1/i^11 = i/(i*i^11)= i/i^12 = i/(i^4)^3 = i/1^3 =i/1 = 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 7673 views around the world You can reuse this answer Creative Commons License