# What is the trigonometric form of  (-2-9i) ?

##### 1 Answer
Dec 18, 2015

$\sqrt{85} {e}^{i \arccos \left(- \frac{2}{85}\right)}$

#### Explanation:

You first need the module of this complex number, given by the formula $\left\mid - 2 - 9 i \right\mid = \sqrt{{\left(- 9\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{85}$.

You can now factorize the complex number by its module : $- 2 - 9 i = \sqrt{85} \left(- \frac{2}{85} - i \frac{9}{85}\right)$. You can now say that $\exists \theta \in \mathbb{R}$ such that $\cos \left(\theta\right) = - \frac{2}{85}$ and $\sin \left(\theta\right) = - \frac{9}{85}$.

So $\theta = \arccos \left(- \frac{2}{85}\right)$.