Answer
$ \sqrt 3 -i$
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 $
Since, $\theta= 330^{\circ}$ is in the Fourth Quadrant , so the reference angle $= 360^{\circ}- 330^{\circ}= 30^{\circ}$
Because the is in Fourth Quadrant, so we will take it as a negative value.
We know that $\sin (-\theta) =- \sin \theta$ and $\cos (-\theta) =\cos \theta$
Therefore, $2 (\cos (-30^{\circ}) +i \sin (-30^{\circ})) = 2(\dfrac{\sqrt 3}{2} -i \dfrac{1}{2} ) \\ = \sqrt 3 -i$
