# How do you write the trigonometric form of sqrt3+i?

Nov 19, 2016

$\sqrt{3} + i = 2 \cos \left(\frac{\pi}{6}\right) + 2 i \sin \left(\frac{\pi}{6}\right)$

#### Explanation:

A number $a + i b$ can be written in trigonometric form as

$r \cos \theta + i r \sin \theta$.

As $r \cos \theta = a$ and $r \sin \theta = b$, squaring and adding them we get ${r}^{2} = {a}^{2} + {b}^{2}$.

As such for $\sqrt{3} + i$, $r = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$

and $\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$

i.e. $\theta = \frac{\pi}{6}$

Hence in trigonometric form

$\sqrt{3} + i = 2 \cos \left(\frac{\pi}{6}\right) + 2 i \sin \left(\frac{\pi}{6}\right)$