How do you convert ( 1 , - sqrt3 )into polar coordinates?

Jan 9, 2016

If $\left(a , b\right)$ is a are the coordinates of a point in Cartesian Plane, $u$ is its magnitude and $\alpha$ is its angle then $\left(a , b\right)$ in Polar Form is written as $\left(u , \alpha\right)$.
Magnitude of a cartesian coordinates $\left(a , b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(1 , - \sqrt{3}\right)$ and $\theta$ be its angle.
Magnitude of $\left(1 , - \sqrt{3}\right) = \sqrt{{\left(1\right)}^{2} + {\left(- \sqrt{3}\right)}^{2}} = \sqrt{1 + 3} = \sqrt{4} = 2 = r$
Angle of $\left(1 , - \sqrt{3}\right) = T a {n}^{-} 1 \left(- \frac{\sqrt{3}}{1}\right) = T a {n}^{-} 1 \left(- \sqrt{3}\right) = - \frac{\pi}{3}$

$\implies$ Angle of $\left(1 , - \sqrt{3}\right) = - \frac{\pi}{3}$

But since the point is in fourth quadrant so we have to add $2 \pi$ which will give us the angle.

$\implies$ Angle of $\left(1 , - \sqrt{3}\right) = - \frac{\pi}{3} + 2 \pi = \frac{- \pi + 6 \pi}{3} = \frac{5 \pi}{3}$

$\implies$ Angle of $\left(1 , - \sqrt{3}\right) = \frac{5 \pi}{3} = \theta$

$\implies \left(1 , - \sqrt{3}\right) = \left(r , \theta\right) = \left(2 , \frac{5 \pi}{3}\right)$
$\implies \left(1 , - \sqrt{3}\right) = \left(2 , \frac{5 \pi}{3}\right)$
Note that the angle is given in radian measure.

Note that the answer $\left(1 , - \sqrt{3}\right) = \left(2 , - \frac{\pi}{3}\right)$ is also correct.