How do you simplify #(3-4i)(-1+2i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer mason m Dec 25, 2015 #5+10i# Explanation: Use the FOIL method of distribution. #(3-4i)(-1+2i)# #=>3(-1)+3(2i)-4i(-1)-4i(2i)# #=>-3+6i+4i-8i^2# #=>-3+10i-8i^2# Now, recall that #i^2=-1#. This is true since #i=sqrt(-1)#. #=>-3+10i-8(-1)# #=>-3+10i+8# #=>5+10i# 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 5930 views around the world You can reuse this answer Creative Commons License