How do you simplify #(6-7i)(2+5i) #? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer sente Dec 16, 2015 #(6-7i)(2+5i) = 47 + 16i# Explanation: #(6-7i)(2+5i) = 6*2 + 6*5i + (-7i)*2 + (-7i)*5i# #= 12 + 30i - 14i - 35i^2# #= 12 + 16i - 35(-1)# #= 12 + 16i + 35# #= 47 + 16i# 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 3619 views around the world You can reuse this answer Creative Commons License