How do you multiply #(4+5i)(4-5i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer A08 Aug 31, 2016 We know that #(a+b)(a-b) = a^2 - b^2# and that # i^2 = -1# It follows #(4 + 5i)(4 - 5i)# #= 4^2 - (5i)^2# #= 16 - (25i^2)# #= 16 + 25# #= 41# 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 6377 views around the world You can reuse this answer Creative Commons License