How do you simplify #i^9-i^-5#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Narad T. Dec 18, 2016 The answer is #=2i# Explanation: We use #i^2=-1# #i^3=-i# #i^4=1# #i^5=i# #i^14=i^(3*4+2)=1*i^2=-1# Therefore, #i^9-i^(-5)=i^9-1/i^5# #=(i^14-1)/i^5# #=(-1-1)/i# #=-(2i)/i^2# #=2i# Answer link Related questions How do I multiply complex numbers? How do I multiply complex numbers in polar form? What is the formula for multiplying complex numbers in trigonometric form? How do I use the modulus and argument to square #(1+i)#? What is the geometric interpretation of multiplying two complex numbers? What is the product of #3+2i# and #1+7i#? How do I use DeMoivre's theorem to solve #z^3-1=0#? How do I find the product of two imaginary numbers? How do you simplify #(2+4i)(2-4i)#? How do you multiply #(-2-8i)(6+7i)#? See all questions in Multiplication of Complex Numbers Impact of this question 3643 views around the world You can reuse this answer Creative Commons License