# How do you simplify (4+ 2i) /( -1 + i)?

May 2, 2018

$\frac{4 + 2 i}{- 1 + i} | \cdot \left(- 1 - i\right)$

$\frac{\left(4 + 2 i\right) \left(- 1 - i\right)}{\left(- 1 + i\right) \left(- 1 - i\right)}$

$\frac{- 2 {i}^{2} - 6 i - 4}{1 - {i}^{2}}$

$\frac{2 - 6 i - 4}{1 + 1}$

$\frac{- 2 - 6 i}{2}$

$= - 1 - 3 i$

#### Explanation:

We want to get rid of $i$ in the the bottom of the fraction in order to get it on Certesian form. We can do this by multiplying with $\left(- 1 - i\right)$.

This will give us,

$\frac{\left(4 + 2 i\right) \left(- 1 - i\right)}{\left(- 1 + i\right) \left(- 1 - i\right)}$

$\frac{- 2 {i}^{2} - 6 i - 4}{1 - {i}^{2}}$

Out from here we know that ${i}^{2} = - 1$ and $- {i}^{2} = 1$. So we can get rid of the ${i}^{2}$ too. Leaving us to

$\frac{- 2 - 6 i}{2}$

$= - 1 - 3 i$