How do you multiply #8i+8(4-2i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer 冠廷 李. Sep 22, 2016 #32-8i# Explanation: #RR byRR# #CCby CC# so #8i+8(4-2i)# =#8i+32-16i# =#32-8i# 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 1366 views around the world You can reuse this answer Creative Commons License