Answer
$$6\left( {\cos {{240}^ \circ } + i\sin {{240}^ \circ }} \right)$$
Work Step by Step
$$\eqalign{
& - 3 - 3i\sqrt 3 \cr
& {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr
& r = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 3\sqrt 3 } \right)}^2}} = 6 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{ - 3\sqrt 3 }}{{ - 3}}} \right) = {60^ \circ } \cr
& {\text{The vector lies is in the Quadrant III, then}} \cr
& \theta = {60^ \circ } + {180^ \circ } = {240^ \circ } \cr
& {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr
& = 6\left( {\cos {{240}^ \circ } + i\sin {{240}^ \circ }} \right) \cr} $$
