Answer
$(\sqrt{3}, \displaystyle \frac{3\pi}{2})$
Work Step by Step
To change from rectangular to polar coordinates, use
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$r^{2}=0^{2}+(-\sqrt{3})^{2}=3$
$r=\pm\sqrt{3}$
$\displaystyle \tan\theta=\frac{-\sqrt{3}}{0}$ is undefined,
so $\theta$ has either the positive y-axis or negative y-axis as its terminal side.
Given the interval for $\theta$, it is either $\displaystyle \frac{\pi}{2}$ or $\displaystyle \frac{3\pi}{2}.$
The point P$(0,-\sqrt{3})$ lies on the negative y-axis,
so we take $\displaystyle \theta=\frac{3\pi}{2}$.
The terminal end of $\theta$ passes through P, so for this angle, r is positive.
Polar coordinates: $(\sqrt{3}, \displaystyle \frac{3\pi}{2})$
