How do you simplify #(6-2i)(1+i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer MeneerNask Feb 17, 2017 You work out the multiplication using FOIL: Explanation: #=6*1+(-2i)*1+6*i+(-2i)*i# #=6-2i+6i-2i^2# Since #i^2=-1#: #=6-2i+6i+2# #=8+4i# 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 2954 views around the world You can reuse this answer Creative Commons License