How do you simplify #(-3 + i)(2 + 7i) #? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Lucio Falabella Jan 22, 2016 #(-3+i)(2+7i)=-13-19i=-(13+19i)# Explanation: Remembering that #i^2=-1# You can think that #(a+ib)# is like a binomial, thus: #(-3+i)(2+7i)=(-3times2-3times7i+2timesi+7i^2)=# #=-6-21i+2i+7times(-1)=(-6-7)+(-21+2)i=-13-19i=-(13+19i)# 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 1975 views around the world You can reuse this answer Creative Commons License