How do you simplify #(-10i)(-9i)-(8i)(6+9i)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer anor277 Oct 10, 2016 #(-10i)xx(-9i)-8i(6+9i)=-18-48i# Explanation: #(-10i)xx(-9i)-8i(6+9i)= (-9xx-10)i^2-48i-72i^2# And so, #(-9xx-10)i^2-48i-72i^2# #=# #18i^2-48i# Of course, we know that #i^2=-1# Thus, #18i^2-48i# #=# #-18-48i# 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 1622 views around the world You can reuse this answer Creative Commons License