How do you simplify #1/(2i)^3#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer anor277 Nov 16, 2016 #1/(2i)^3=-1/(8i)# Explanation: #1/(2i)^3=1/(2^3i^3)=1/8xx1/i^3=1/8xx1/i^2xx1/i=1/8xx(1/-1)xx1/i# #=-1/(8i)# 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 4050 views around the world You can reuse this answer Creative Commons License