How do you multiply #(4-2i) (3 + 2i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Binayaka C. Apr 19, 2018 #16 +2 i # Explanation: #(4-2 i)(3+2 i) = 12 + 8 i - 6 i - 4 i^2 # #=12 + 2 i - 4 * (-1) ; [i^2 = -1] # #=12 + 2 i +4 = 16 +2 i # [Ans] 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 1287 views around the world You can reuse this answer Creative Commons License