How do you simplify #(1-7i)^2#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Noah G Oct 25, 2016 #=(1 - 7i)(1 - 7i)# #=1 - 7i - 7i + 49i^2# We know that #i^2 = -1#, so: #=1 - 14i + 49(-1)# #=1 - 14i - 49# #=-48 - 14i# Hopefully this helps! 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 6086 views around the world You can reuse this answer Creative Commons License