# How do you divide ( -i-8) / (-i +7 ) in trigonometric form?

Dec 17, 2015

$\frac{- i - 8}{- i + 7} = \sqrt{\frac{65}{50}} {e}^{\arccos \left(- \frac{8}{\sqrt{65}}\right) - \arccos \left(- \frac{7}{\sqrt{50}}\right)}$

#### Explanation:

Usually I always simplify this kind of fraction by using the formula $\frac{1}{z} = \frac{z \overline{z}}{\left\mid z \right\mid} ^ 2$ so I'm not sure what I'm going to tell you works but this is how I'd solve the problem if I only wanted to use trigonometric form.

$\left\mid - i - 8 \right\mid = \sqrt{64 + 1} = \sqrt{65}$ and $\left\mid - i + 7 \right\mid = \sqrt{50}$. Hence the following results : $- i - 8 = \sqrt{65} \left(- \frac{8}{\sqrt{65}} - \frac{i}{\sqrt{65}}\right)$ and $- i + 7 = \sqrt{50} \left(\frac{7}{\sqrt{50}} - \frac{i}{\sqrt{50}}\right)$

You can find $\alpha , \beta \in \mathbb{R}$ such that $\cos \left(\alpha\right) = - \frac{8}{\sqrt{65}}$, $\sin \left(\alpha\right) = - \frac{1}{\sqrt{65}}$, $\cos \left(\beta\right) = \frac{7}{\sqrt{50}}$ and $\sin \left(\beta\right) = - \frac{1}{\sqrt{50}}$.

So $\alpha = \arccos \left(- \frac{8}{\sqrt{65}}\right) = \arcsin \left(- \frac{1}{\sqrt{65}}\right)$ and $\beta = \arccos \left(- \frac{7}{\sqrt{50}}\right) = \arcsin \left(- \frac{1}{\sqrt{50}}\right)$, and we can now say that $- i - 8 = \sqrt{65} {e}^{\arccos} \left(- \frac{8}{\sqrt{65}}\right)$ and $- i + 7 = \sqrt{50} {e}^{\arccos} \left(- \frac{7}{\sqrt{50}}\right)$.