How do you divide (3+9i)/(-6-6i)?

Jul 25, 2016

$- 1 - \frac{i}{2}$.

Explanation:

Let $z = \frac{3 + 9 i}{- 6 - 6 i}$

$\therefore z = \frac{3 \left(1 + 3 i\right)}{- 6 \left(1 + i\right)} = - \frac{1}{2} \left(\frac{1 + 3 i}{1 + i}\right)$

$\therefore z = - \frac{1}{2} \left(\frac{1 + 3 i}{1 + i}\right) \left(\frac{1 - i}{1 - i}\right)$

$\therefore z = - \frac{1}{2} \left(\frac{1 - i + 3 i - 3 {i}^{2}}{{1}^{2} - {i}^{2}}\right)$

$\therefore z = - \frac{1}{2} \frac{\left(1 + 2 i - 3 \left(- 1\right)\right)}{1 - \left(- 1\right)}$

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

$\therefore z = - 1 - \frac{i}{2}$.

Jul 25, 2016

$- 1 - \frac{1}{2} i$

Explanation:

You can multiply numerator and denominator by the expression

$- 6 + 6 i$:

$\frac{\left(3 + 9 i\right) \left(- 6 + 6 i\right)}{\left(- 6 - 6 i\right) \left(- 6 + 6 i\right)} =$

$\frac{- 18 + 18 i - 54 i + 54 {i}^{2}}{36 - 36 {i}^{2}} =$

Since ${i}^{2} = - 1$, you will have:

$\frac{- 18 - 36 i - 54}{36 + 36} =$

$\frac{- 72 - 36 i}{72} =$

$- 1 - \frac{1}{2} i$