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

Feb 17, 2016

13

#### Explanation:

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

$\left(3 + 2 i\right) \left(3 - 2 i\right) = 9 - 6 i + 6 i - 4 {i}^{2}$

[note : ${i}^{2} = {\left(\sqrt{-} 1\right)}^{2} = - 1$]

hence 9 - 6i + 6i $- 4 {i}^{2} = 9 + 4 = 13$

[ 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.