Answer
$$2\left( {\cos {{330}^ \circ } + i\sin {{330}^ \circ }} \right)$$
Work Step by Step
$$\eqalign{
& \sqrt 3 - i \cr
& {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr
& r = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2}} = 2 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right) = - {30^ \circ } \cr
& {\text{The vector lies is in the Quadrant IV, then}} \cr
& \theta = - {30^ \circ } + {360^ \circ } = {330^ \circ } \cr
& {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr
& = 2\left( {\cos {{330}^ \circ } + i\sin {{330}^ \circ }} \right) \cr} $$
