Answer
$$\sqrt {13} \left( {\cos {{56.31}^ \circ } + i\sin {{56.31}^ \circ }} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Rectangular Form }}2 + 3i \cr
& {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr
& r = \sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} = \sqrt {13} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{3}{2}} \right) \cr
& \theta \approx {56.31^ \circ } \cr
& {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr
& = \sqrt {13} \left( {\cos {{56.31}^ \circ } + i\sin {{56.31}^ \circ }} \right) \cr} $$
