Question

how do I solve (3+i)(3-i)

Answer 1

$(3+i)(3-i)=3(3-i)+i(3-i)=9-3i+3i-i^2=9-(-1)=\\\\=9+1=10$

Answer 2

(3+I)(3-1)
-i^2+9=0
-I^2=-9
I^2=9
i= + or - 3