How do you simplify #(-3 - 2i)(-3 + 3i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Alan P. Feb 5, 2016 #15-3i# Explanation: #{: (xx," | ",-3,-2i), ("------",,"------","------"), (-3," | ",+9,+6i), (+3i," | ",-9i,-6i^2), (,"------","------","------"), (,9,-3i,-6i^2) :}# Since #i^2=-1# #rarr -6i^2= +6# Therefore #(-3-2i)(-3+3i)=9-3i-6i^2 = color(green)(15-3i)# 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 1994 views around the world You can reuse this answer Creative Commons License