# How do you find the product of each complex number 8-2i and its conjugate?

Dec 17, 2015

I found $68$

#### Explanation:

The conjugade is: $8 + 2 i$
so you have:
$\left(8 - 2 i\right) \left(8 + 2 i\right) =$ use the distributive property:
$= \left(8 \cdot 8\right) + \left(2 \cdot 8\right) i - \left(2 \cdot 8\right) i - \left(2 \cdot 2\right) \left(i \cdot i\right) =$
$= 64 + \cancel{\left(2 \cdot 8\right) i} \cancel{- \left(2 \cdot 8\right) i} - 4 {i}^{2} =$
but:
${i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1$
so:
$64 + 4 = 68$ a Pure Real!!!