Answer
$\dfrac{-3 \sqrt {3}}{2}+i \dfrac{3}{2}$
Work Step by Step
The given complex number can be written in the rectangular form as follows: $r (\cos \theta +i \sin \theta) = a+i b$
Where, $a=r \cos \theta ; b =r \sin \theta $
Let us suppose that $z =3(\cos 150^{\circ} +i \sin 150^{\circ} ) ....(1) $
Now, $\cos 150^{\circ}=\cos (90^{\circ}+60^{\circ}) =-\sin 60^{\circ}=\dfrac{-\sqrt {3}}{2}$ and $\sin 150^{\circ}=\sin (90^{\circ}+60^{\circ}) =\cos 60^{\circ}=\dfrac{1}{2}$
Therefore, the equation (1) becomes:
$z= 3(\dfrac{-\sqrt {3}}{2} +i \dfrac{1}{2})=\dfrac{-3 \sqrt {3}}{2}+i \dfrac{3}{2}$
