# How do you convert  sqrt(3)+i to polar form?

Jun 23, 2016

I found: $2 \angle {30}^{\circ}$

#### Explanation:

Have a look:

Jun 23, 2016

$\left(2 , \frac{\pi}{6}\right)$

#### Explanation:

To convert cartesian coordinates (x ,y) to Polar coordinates $\left(r , \theta\right)$
Use the formulae which link them.

•r=sqrt(x^2+y^2)

and since $\sqrt{3} + i = \left(\sqrt{3} , 1\right)$ is in the 1st quadrant , then.

•theta=tan^-1(y/x)

here x =$\sqrt{3} \text{ and } y = 1$

$\Rightarrow r = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}} = \sqrt{4} = 2$

and $\theta = {\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \text{ or } {30}^{\circ}$

$\Rightarrow \left(\sqrt{3} , 1\right) \to \left(2 , \frac{\pi}{6}\right)$