# What is the polar form of ( -4,5 )?

Dec 17, 2015

The polar form of (-4,5) has $\sqrt{41}$ as module and $\arccos \left(- \frac{4}{\sqrt{41}}\right)$ as argument.

#### Explanation:

You can use Pythagoras theorem or the complex numbers. I'm gonna use the complex numbers because it is simpler to write down and to explain as I always do that and english is not my mother language.

By identifying ${\mathbb{R}}^{2}$ as the complex plan $\mathbb{C}$, $\left(- 4 , 5\right)$ is the complex number $- 4 + 5 i$. Its module is $\left\mid - 4 + 5 i \right\mid = \sqrt{{5}^{2} + {\left(- 4\right)}^{2}} = \sqrt{41}$.

We now need the argument of this complex number. We know its module, so we can write that $- 4 + 5 i = \sqrt{41} \left(- \frac{4}{\sqrt{41}} + i \frac{5}{\sqrt{41}}\right)$.

We know that when we factorize by the module, we get the cosine and the sine of a real number. It means that $\exists \alpha \in \mathbb{R}$ such that $\cos \left(\alpha\right) = - \frac{4}{\sqrt{41}}$ and $\sin \left(\alpha\right) = \frac{5}{\sqrt{41}}$. So $\alpha = \arccos \left(- \frac{4}{\sqrt{41}}\right)$ which is the argument of (-4,5).