Answer
$$\left( {2,\frac{\pi }{3}} \right)$$ or $$\left( { - 2,\frac{{4\pi }}{3}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {1,\sqrt 3 } \right) \Rightarrow x = - 1{\text{ and }}y = 0 \cr
& {\text{Converting to cartesian coordinates }}\left( {r,\theta } \right).{\text{ Where}} \cr
& r = \sqrt {{x^2} + {y^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) \cr
& {\text{We obtain}}{\text{,}} \cr
& r = \sqrt {{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} = 2 \cr
& {\text{Coordinate }}x{\text{ }}\left( {{\text{positive}}} \right){\text{, Coordinate }}y{\text{ }}\left( {{\text{positive}}} \right),{\text{ Quadrant I}} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \frac{\pi }{3} \cr
& {\text{The polar coordinates are:}} \cr
& \left( {r,\theta } \right) = \left( {2,\frac{\pi }{3}} \right) \cr
& or \cr
& \left( { - r,\theta + \pi } \right) = \left( { - 2,\frac{{4\pi }}{3}} \right) \cr} $$
