How do you simplify #(2i)(-3+5i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Dwight Jan 16, 2017 The answer is #-10-6i# Explanation: I like to think of treating the #i# just like it was a variable. Multiply the #2i# term through the bracket in the same way you would if it was #x# #-6i+10i^2# #i^2# equals -1, so the equation becomes #-6i-10# which would usually be written #-10-6i# 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 2781 views around the world You can reuse this answer Creative Commons License