Answer
$-2- i \ 2 \sqrt 3$
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, $\theta= 240^{\circ}$ is in the Third Quadrant , so the reference angle $= 240^{\circ}-180^{\circ}=60^{\circ}$
Therefore, $4 (\cos 240^{\circ} +i \sin 240^{\circ}) = 4(- \dfrac{1}{2}-i \dfrac{\sqrt 3}{2} ) \\ =-2- i \ 2 \sqrt 3$
