How do you simplify #(5-3i)^2#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 2 Answers Harish Chandra Rajpoot Jul 28, 2018 #16-30i# Explanation: Given that #(5-3i)^2# #=(5-3i)(5-3i)# #=25-15i-15i+9i^2# #=25-30i-9# #=16-30i# Answer link Kalyanam S. Jul 28, 2018 #color(indigo)(=> 16 - 30 i# Explanation: #(a - b)^2 = a^2 - 2ab + b^2#, identity. #(5- 3 i)^2 = 5^2 - (2*5*3 i) + (3 i)^2# #i^2 = -1# #=> 25 - 30 i - 9# #=> 25 - 9 - 30 i# #color(indigo)(=> 16 - 30 i# 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 13397 views around the world You can reuse this answer Creative Commons License