# How do you simplify (8 - 6i)/(3i)?

Feb 7, 2016

$- 2 - \frac{8}{3} i$

#### Explanation:

require the denominator to be real. To achieve this multiply numerator and denominator by 3i

$3 i \times 3 i = 9 {i}^{2}$

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

$\Rightarrow \frac{8 - 6 i}{3 i} = \frac{3 i \left(8 - 6 i\right)}{3 i .3 i} = \frac{24 i - 18 {i}^{2}}{9 {i}^{2}}$

$= \frac{24 i + 18}{-} 9 = \frac{18}{-} 9 + \frac{24 i}{-} 9 = - 2 - \frac{8}{3} i$

Feb 7, 2016

This one is reasonably simple: multiply top and bottom by $- \frac{i}{3}$. Doing this and rearranging leads to the simplified answer $- 2 - \frac{8}{3} i$.

#### Explanation:

Since there is a single complex number - and only the imaginary part of it - on the bottom row, I decided to multiply top and bottom of the fraction by a number that will yield a bottom line of 1. That number is $- \frac{i}{3}$. Multiplying by ($- \frac{i}{3}$)/( $- \frac{i}{3}$) is the same thing as multiplying by 1, which means the simplified number we end up with is the same number we started with.

$\frac{8 - 6 i}{3 i} \cdot \frac{- \frac{i}{3}}{- \frac{i}{3}} = \frac{- \frac{8}{3} i - 2}{1}$

Written in more traditional form, this is $- 2 - \frac{8}{3} i$.