Нужна помощь¡!!!!!!!!!!!!!

Представим число в тригонометрической форме:
$z=1-i\sqrt{3}=2(\frac{1}{2}-i \frac{\sqrt{3}}{2})=2(cos (-\frac{\pi}{3})+ isin (-\frac{\pi}{3}))$
$\sqrt[3]{z} = \sqrt[3]{2}(cos(\frac{\frac{-\pi}{3}+2\pi k}{3})+ isin(\frac{\frac{-\pi}{3}+2\pi k}{3})),$ $k=0,1,2$
$x_{1}=\sqrt[3]{2}(cos(\frac{\frac{-\pi}{3}}{3})+ isin(\frac{\frac{-\pi}{3}}{3}))=$ $\sqrt[3]{2}(cos(\frac{ -\pi }{9})+ isin(\frac{ -\pi }{9})), k=0$
$x_{2}=\sqrt[3]{2}(cos(\frac{\frac{-\pi}{3}+2 \pi }{3})+ isin(\frac{\frac{-\pi}{3}+2 \pi }{3}))=$ $\sqrt[3]{2}(cos(\frac{ 5\pi }{9})+ isin(\frac{ 5\pi }{9})), k=1$
$x_{3}=\sqrt[3]{2}(cos(\frac{\frac{-\pi}{3}+4 \pi }{3})+ isin(\frac{\frac{-\pi}{3}+4 \pi }{3}))=$ $\sqrt[3]{2}(cos(\frac{ 11\pi }{9})+ isin(\frac{ 11\pi }{9})), k=2$