Answer
$$12\left( {\cos {{90}^ \circ } + i\sin {{90}^ \circ }} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Rectangular Form }}12i \cr
& {\text{Use }}r = \sqrt {{a^2} + {b^2}} {\text{ and }}\theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right),{\text{ so}} \cr
& r = \sqrt {{{\left( 0 \right)}^2} + {{\left( {12} \right)}^2}} = 12 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{12}}{0}} \right) = {90^ \circ } \cr
& {\text{write the vector in the trigonometric form }}r\left( {\cos \theta + i\sin \theta } \right) \cr
& = 12\left( {\cos {{90}^ \circ } + i\sin {{90}^ \circ }} \right) \cr} $$