Answer
$2cis240^\circ$ or $2(cos 240^\circ + isin240^\circ)$
Work Step by Step
$arctan(\frac{-\sqrt{3}}{-1}) = 60^\circ$, but since, $- 1 - i\sqrt{3}$ is in the 3rd quadrant, its angle is $240^\circ$.
The absolute value of $- 1 - i\sqrt{3}$ is $\sqrt{(-1)^2 + (-\sqrt{3})^2} = 2$, the equivalent trigonometric form is $2cis240^\circ$ or $2(cos 240^\circ + isin240^\circ)$.
