Answer
$$\sqrt {17} \left( {\cos {{165.96}^ \circ } + i\sin {{165.96}^ \circ }} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Rectangular Form }} - 4 + i \cr
& {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr
& r = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {17} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{1}{{ - 4}}} \right) \cr
& {\text{The vector lies is in the Quadrant II, then}} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{1}{{ - 4}}} \right) + {180^ \circ } \cr
& \theta \approx {165.96^ \circ } \cr
& {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr
& = \sqrt {17} \left( {\cos {{165.96}^ \circ } + i\sin {{165.96}^ \circ }} \right) \cr} $$
