How do you simplify #(-9 - sqrt(-3)) ^2#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Shwetank Mauria Jul 16, 2016 #(-9-sqrt(-3))^2=78+18sqrt(-3)# Explanation: As #(a-b)^2=a^2-2ab+b^2# #(-9-sqrt(-3))^2# = #(-9)^2-2×(-9)×sqrt(-3)+(sqrt(-3))^2# or = #81+18sqrt(-3)-3# or = #78+18sqrt(-3)# 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 1422 views around the world You can reuse this answer Creative Commons License