Answer
The polar coordinates of $\left( -1,-\sqrt{3} \right)$ are $\left( 2,\frac{4\pi }{3} \right)$.
Work Step by Step
The polar coordinates of the point are $\left( r,\theta \right)$.
Now, rewrite the polar coordinates in terms of rectangular coordinates as below:
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1)
$\tan \theta =\frac{y}{x}$ …… (2)
Substituting the values of $x\ \text{ and }\ y$ in (1) and (2), we get
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -\sqrt{3} \right)}^{2}}} \\
& =\sqrt{1+3}=\sqrt{4} \\
& r=2
\end{align}$
And,
$\begin{align}
& \tan \theta =\frac{y}{x} \\
& =\frac{-\sqrt{3}}{-1} \\
& \tan \theta =\sqrt{3}
\end{align}$
Hence,
$\tan \theta =\sqrt{3}$
And,
$\tan \frac{\pi }{3}=\sqrt{3}$
Also, $\theta $ lies in the third quadrant which means
$\begin{align}
& \theta =\pi +\frac{\pi }{3} \\
& \theta =\frac{4\pi }{3}
\end{align}$
Therefore, the polar coordinates of $\left( -1,-\sqrt{3} \right)$ are $\left( 2,\frac{4\pi }{3} \right)$.
