# When -3-2i is multiplied by its conjugate, what is the result?

Dec 17, 2016

The result is $13$.

#### Explanation:

A binomial's conjugate is the same expression but with the opposite sign on the 2nd term. For example: $\left(a + b\right)$ and $\left(a - b\right)$ are conjugates. This is useful when we have a fraction with a radical (or imaginary) binomial for a denominator, and we wish to "rationalize" it to remove that.

For any complex number $a + b i$, the complex conjugate is $a - b i$. Again, just the same binomial with the imaginary part negated. When we multiply a complex number $a + b i$ by its conjugate $a - b i$, the result (by FOIL-ing) is

$\left(a + b i\right) \left(a - b i\right) = {a}^{2} - \left(a b\right) i + \left(a b\right) i - {b}^{2} {i}^{2}$
$\textcolor{w h i t e}{\left(a + b i\right) \left(a - b i\right)} = {a}^{2} - \cancel{\left(a b\right) i} + \cancel{\left(a b\right) i} - {b}^{2} \left(\text{-} 1\right)$
$\textcolor{w h i t e}{\left(a + b i\right) \left(a - b i\right)} = {a}^{2} + {b}^{2}$

See what happened? The imaginary part is no longer there. Multiplying a complex number by its conjugate turns it into a completely real number.

In this example, we're asked to multiply $\text{-} 3 - 2 i$ by its conjugate (which is $\text{-} 3 + 2 i$). The result is

$\left(\text{-"3-2i)("-} 3 + 2 i\right) = 9 - 6 i + 6 i - 4 {i}^{2}$
color(white)(("-"3-2i)("-"3+2i))=9-cancel(6i)+cancel(6i)-4("-"1)
$\textcolor{w h i t e}{\left(\text{-"3-2i)("-} 3 + 2 i\right)} = 9 + 4$
$\textcolor{w h i t e}{\left(\text{-"3-2i)("-} 3 + 2 i\right)} = 13$

## Bonus:

The quick trick to find the product of a complex number and its conjugate: square each coefficient, and then add the squares. That'll be your result. (That's actually right from the general formula above.)