$32 (\cos 60^{\circ} + i \sin 60^{\circ})$
Next, using DeMoivre's find the 5 th root of
$\begin{matrix} 16 + 16 \sqrt{3} i = \\ 32 (\cos 60^{\circ} + i \sin {6.0}^{\circ}) \end{matrix}$
Find $r$ and $θ$ where $θ$ is in the third quadrant.
$\begin{matrix} r^{5} = 32 \\ 5 θ = 60^{\circ}, 420^{\circ}, 780^{\circ}, 1140^{\circ}, 1500^{\circ} \\ r = 2 θ = [?] \circ \end{matrix}$
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Answer

$θ = 240^{\circ}$

Steps

Step 1 ： $r = 2$

Step 2 ： $5 θ = 60^{\circ}, 420^{\circ}, 780^{\circ}, 1140^{\circ}, 1500^{\circ}$

Step 3 ： $θ = 240^{\circ}$