How do you multiply #(3+2i)^2#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer anor277 Jan 16, 2017 You remember that #i^2=-1# Explanation: #(3+2i)^2=(3+2i)(3+2i)# #9+6i+6i+4i^2# #9+12i+4(-1)# #9+12i-4=5+12i# What would #(3+2i)(3-2i)=??# 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 14240 views around the world You can reuse this answer Creative Commons License