# How do you convert 1+jsqrt3 to polar form?

Jul 23, 2016

$1 + j \sqrt{3} = 2 \left(\cos \left(\frac{\pi}{3}\right) + j \sin \left(\frac{\pi}{3}\right)\right)$

#### Explanation:

To write the complex number, say $a + j b$, where ${j}^{2} = - 1$ in polar coordinates, we write them as

$z = r \left(\cos \theta + j \sin \theta\right)$.

Now as $r \cos \theta = a$ and $r \sin \theta = b$.

$r = \sqrt{{a}^{2} + {b}^{2}}$, $\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

Hence here for complex number $1 + j \sqrt{3}$

$r = \sqrt{{1}^{2} + {\left(\sqrt{5}\right)}^{2}} = \sqrt{4} = 2$ and $\theta = {\tan}^{- 1} \left(\frac{\sqrt{3}}{1}\right) = {\tan}^{- 1} \sqrt{3} = \frac{\pi}{3}$

Hence $1 + j \sqrt{3} = 2 \left(\cos \left(\frac{\pi}{3}\right) + j \sin \left(\frac{\pi}{3}\right)\right)$