# How do you divide 20 / (3+i)?

Mar 16, 2018

$6 - 2 i$

#### Explanation:

Let's turn $i$ to a square root.

$\implies \frac{20}{3 + \sqrt{- 1}}$ Let's multiply this by the conjugate of $3 + \sqrt{- 1}$

Remember that the conjugate of $a + b i$ is $a - b i$

We have:

$\implies \frac{20}{3 + \sqrt{- 1}} \cdot \frac{3 - \sqrt{- 1}}{3 - \sqrt{- 1}}$

$\implies \frac{20 \left(3 - \sqrt{- 1}\right)}{\left(3 - \sqrt{- 1}\right) \left(3 + \sqrt{- 1}\right)}$

Expand this to get:

$\implies \frac{60 - 20 \sqrt{- 1}}{9 + 3 \sqrt{- 1} - 3 \sqrt{- 1} - {\left(\sqrt{- 1}\right)}^{2}}$

$\implies \frac{60 - 20 \sqrt{- 1}}{9 - {\left(\sqrt{- 1}\right)}^{2}}$

$\implies \frac{60 - 20 \sqrt{- 1}}{9 - \left(- 1\right)}$

$\implies \frac{60 - 20 \sqrt{- 1}}{10}$ Simplify to get:

$\implies 6 - 2 \sqrt{- 1}$

$\implies 6 - 2 i$