How do you simplify #(8-2i)^2#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer GiĆ³ Jul 26, 2016 I found: #60-32i# Explanation: We can rite it as: #(8-2i)(8-2i)=(8*8)-(2*8i)-(2*8i)+(2*2*i*i)=64-16i-16i+4i^2=# but #i^2=(sqrt(-1))^2=-1# so: #=64-32i-4=60-32i# 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 3141 views around the world You can reuse this answer Creative Commons License