How do you simplify #(4-2i) (3 + 2i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Shwetank Mauria Jun 14, 2016 #(4-2i)(3+2i)=16+2i# Explanation: To simplify #(4-2i)(3+2i)#, just multiply them like two binomials, but remember #i^2=-1#. Hence, #(4-2i)(3+2i)# = #4(3+2i)-2i(3+2i)# = #12+8i-6i-4i^2# = #12+2i-4(-1)# = #12+2i+4# = #16+2i# 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 4238 views around the world You can reuse this answer Creative Commons License