How do you multiply #(7i)(6i)(-3-5i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Henry W. Oct 8, 2016 #126+210i# Explanation: Multiplying #7i*6i# we get #42i^2=-42# (recall that #i^2=-1#) Now multiply #-42(-3-5i)#, which equals #126+210i# 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 2535 views around the world You can reuse this answer Creative Commons License