How do you multiply #-3i(9-6i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Alan N. Aug 14, 2016 #-9(2+3i)# Explanation: #-3i(9-6i) = -27i+18i^2# Since #i^2 = -1# #-27i+18i^2 = -27i-18# #=-9(2+3i)# 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 1356 views around the world You can reuse this answer Creative Commons License