Como você simplifica # (3 + 2i) (3-2i) #?

distribute the brackets using FOIL (or any method you use ).

#(3 + 2i )(3 - 2i ) = 9 - 6i + 6i - 4i^2#

[note : #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.