# How do you simplify #(3+2i)(3-2i)#?

##### 1 Answer

Feb 17, 2016

13

#### Explanation:

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.