How do you multiply #(2-4i)(3+5i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Steve M Nov 12, 2016 # (2-4i)(3+5i) =26-2i # Explanation: # (2-4i)(3+5i) =(2)(3)+(2)(5i)+(-4i)(3)+(-4i)(5i) # # :. (2-4i)(3+5i) =6+10i-12i-20i^2 # # :. (2-4i)(3+5i) =6-2i-20i^2 # Now #i^2=-1#, so # :. (2-4i)(3+5i) =6-2i+20 # # :. (2-4i)(3+5i) =6-2i+20 # # :. (2-4i)(3+5i) =26-2i # 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 4463 views around the world You can reuse this answer Creative Commons License