If [tex]z=-1+i \sqrt{3} [/tex], then
[tex]Rez=-1[/tex]
[tex]Imz= \sqrt{3} [/tex] and
[tex]|z|= \sqrt{Re^2z+Im^2z}= \sqrt{1^2+ (\sqrt{3})^2 }= \sqrt{4}=2 [/tex].
Then [tex]cos\theta = \frac{Rez}{|z|} = \frac{-1}{2} [/tex][tex]sin\theta = \frac{Imz}{|z|}= \frac{ \sqrt{3} }{2} [/tex].
Since [tex]\theta \in (-\pi,\pi][/tex], you can conclude that [tex]\theta= \frac{2\pi}{3} [/tex].
The triginometric form of z is [tex]z=|z|(cos\theta+isin\theta)=2(cos \frac{2\pi}{3} +isin \frac{2\pi}{3} )[/tex].
Use
formula [tex] \sqrt[3]{z} =\{ \sqrt[3]{|z|} (cos \frac{\theta+2\pi
k}{n}+isin \frac{\theta+2\pi k}{n} ), k=0,1,2\}[/tex] to find cube
root.
For k=0, [tex]z_1= \sqrt[3]{2} (cos \frac{
\frac{2\pi}{3} }{3} +isin\frac{ \frac{2\pi}{3} }{3})= \sqrt[3]{2} (cos
40^0+isin 40^0)[/tex],
For k=1, [tex]z_1= \sqrt[3]{2} (cos
\frac{ \frac{2\pi}{3}+2\pi }{3} +isin\frac{ \frac{2\pi}{3}+2\pi }{3})=
\sqrt[3]{2} (cos 160^0+isin 160^0)[/tex],
For k=2,
[tex]z_1= \sqrt[3]{2} (cos \frac{ \frac{2\pi}{3}+4\pi }{3} +isin\frac{
\frac{2\pi}{3}+4\pi }{3})= \sqrt[3]{2} (cos 280^0+isin 280^0)[/tex].
Answer: The correct choice is C.
