How do you multiply #(3-2i)(5-6i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Karthik G Jan 14, 2016 Steps are given below. Explanation: #(3-2i)(5-6i)# Use FOIL #=(3)(5) + (3)(-6i) - (2i)(5) + (-2i)(-6i)# #=15 - 18i - 10i + 12i^2# #color(red) "note : i² =-1"# #=15 -28i + 12(-1)# #=15-28i-12# #=3-28i# 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 4439 views around the world You can reuse this answer Creative Commons License