# How do you divide (-2-5i)/(3i)?

Feb 13, 2016

Solution is $- \frac{5}{3} + \left(\frac{2}{3}\right) i$

#### Explanation:

To solve (−2−5i)/(3i)

multiply numerator and denominator by $i$ (note ${i}^{2} = - 1$)

and we get

(i*(−2−5i))/(i*(3i) or $\frac{- 2 i - 5 {i}^{2}}{3 {i}^{2}}$

or $\frac{- 2 i - 5 \left(- 1\right)}{3 \cdot - 1}$ i.e. $\frac{5 - 2 i}{-} 3$, which is equivalent to

$- \frac{5}{3} + \left(\frac{2}{3}\right) i$

Feb 13, 2016

$\frac{2}{3} i - \frac{5}{3}$

#### Explanation:

The denominator of the fraction is required to be real. To achieve this multiply the numerator and denominator by 3i.

$\frac{- 2 - 5 i}{3 i} \times \frac{3 i}{3 i}$

$= \frac{- 6 i - 15 {i}^{2}}{9 {i}^{2}}$

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

$= \frac{- 6 i + 15}{-} 9 = \frac{2}{3} i - \frac{5}{3}$