# How do you divide 4/(-3-6i)?

Aug 30, 2016

$- \frac{4}{15} + \frac{8}{15} i$

#### Explanation:

To divide this fraction we require the denominator to be $\textcolor{b l u e}{\text{real}}$

To obtain this multiply (-3 - 6i) by it's $\textcolor{b l u e}{\text{conjugate}}$

The conjugate of (-3 - 6i) is (-3 + 6i). Note that the real part remains unchanged while the $\textcolor{red}{\text{sign}}$ of the imaginary part is reversed.

$\Rightarrow \left(- 3 - 6 i\right) \left(- 3 + 6 i\right) = 9 - 18 i + 18 i - 36 {i}^{2} = 45 \text{ a real number}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Since this is a fraction, we also have to multiply the numerator by (-3 + 6i)

$\frac{4}{- 3 - 6 i} \times \frac{- 3 + 6 i}{- 3 + 6 i} = \frac{4 \left(- 3 + 6 i\right)}{\left(- 3 - 6 i\right) \left(- 3 + 6 i\right)}$

$= \frac{- 12 + 24 i}{45} = - \frac{12}{45} + \frac{24}{45} i = - \frac{4}{15} + \frac{8}{15} i$