How do you simplify #(2-2i)^2#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer maganbhai P. Aug 10, 2018 #z=0-8i# Explanation: Let , #z=(2-2i)^2 ,where ,z inCC# #:.z=4-8i+4i^2# #:.z=4-8i+4(-1)to[because i^2=-1]# #:.z=4-8i-4# #:.z=0-8i# 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 9486 views around the world You can reuse this answer Creative Commons License