# How do you convert  -6i  to polar form?

May 22, 2016

Polar form of $- 6 i$ is $\left(6 , - \frac{\pi}{2}\right)$

#### Explanation:

A complex number $a + i b$ in polar form is written as

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

Hence $r = \sqrt{{a}^{2} + {b}^{2}}$

As $- 6 i$ can be written as $0 - 6 i$

r=sqrt()^2+(-6)^2)=sqrt36=6 and hence

$\cos \theta = \frac{0}{6} = 0$ and $\sin \theta = - \frac{6}{6} = - 1$

Hence, $\theta = - \frac{\pi}{2}$ and

Polar form of $- 6 i$ is $\left(6 , - \frac{\pi}{2}\right)$