How do you simplify #(3+2i)(3-2i)#?
distribute the brackets using FOIL (or any method you use ).
#(3 + 2i )(3 - 2i ) = 9 - 6i + 6i - 4i^2#
#i^2 =(sqrt-1)^2 = - 1 #]
hence 9 - 6i + 6i
# - 4i^2 = 9 + 4 = 13#
[ note that the result is a real number. This is a useful 'tool' in complex numbers]
If a + bi is a complex number then a - bi is called the 'conjugate'
and the product (a + bi)(a - bi) is always real.