Calculate the argument of $ (-2{\sqrt 3} - 2i)^5$

Question

So I'm supposed to calcualte the argument and the answer is supposed to be $\frac{11}{6}π$.

Instead of that I'm getting $\frac{5}{6}π$ because $\frac{-2}{-2{\sqrt 3}} = \frac{1}{{\sqrt 3}} $ the argument of that being $\frac{π}{{6}}$ and when I multiply it by $5,$ I get $\frac{5}{6}π$

where is my mistake?

Answer 1

$$arg((-2-2\sqrt{3})^5) = arg((-1)^5)+arg((2+2\sqrt{3})^5)$$ $$=arg(-1)+arg((2+2\sqrt{3})^5)$$ $$=\pi+5arg(2+2\sqrt{3}) = \pi+5\pi/6 = 11\pi/6.$$

Answer 2

$$(-2\sqrt3-2i)^5=\left(4\left(-\frac{\sqrt3}{2}-\frac{1}{2}i\right)\right)^5=$$ $$=1024(\cos(210^{\circ}\cdot5)+i\sin(210^{\circ}\cdot5))=$$ $$=1024(\cos330^{\circ}+i\sin330^{\circ}),$$ which gives the answer: $$330^{\circ}.$$