Answer
$$2\sqrt 2 \left( {\cos {{135}^ \circ } + i\sin {{135}^ \circ }} \right)$$
Work Step by Step
$$\eqalign{
& - 2 + 2i \cr
& {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr
& r = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2}} = 2\sqrt 2 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{2}{{ - 2}}} \right) = - {45^ \circ } \cr
& {\text{The vector lies is in the Quadrant II, then}} \cr
& \theta = - {45^ \circ } + {180^ \circ } = {135^ \circ } \cr
& {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr
& = 2\sqrt 2 \left( {\cos {{135}^ \circ } + i\sin {{135}^ \circ }} \right) \cr} $$
