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

May 4, 2016

$\sqrt{3} + i$ in polar form is written as $2 \cos \left(\frac{\pi}{6}\right) + 2 i \sin \left(\frac{\pi}{6}\right)$ or 2e^(i(pi/6)

#### Explanation:

Complex number $a + i b$ is written as $r \cos \theta + i r \sin \theta$ or $r {e}^{i \theta}$ in polar form.

where, $r = \sqrt{{a}^{2} + {b}^{2}} 3 \mathmr{and}$tantheta=b/a$\mathmr{and}$theta=arctan(b/a)

Hence for $\sqrt{3} + i$

$r = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}} = \sqrt{4} = 2$ and $\theta = \arctan \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

Hence $\sqrt{3} + i$ in polar form is written as $2 \cos \left(\frac{\pi}{6}\right) + 2 i \sin \left(\frac{\pi}{6}\right)$

or 2e^(i(pi/6)#