Question

Which expression is a cube root of -1+i sqrt3

Answer 1

If $z=-1+i \sqrt{3}$, then 

$Rez=-1$
$Imz= \sqrt{3}$ and 
$|z|= \sqrt{Re^2z+Im^2z}= \sqrt{1^2+ (\sqrt{3})^2 }= \sqrt{4}=2$.
Then $cos\theta = \frac{Rez}{|z|} = \frac{-1}{2}$$sin\theta = \frac{Imz}{|z|}= \frac{ \sqrt{3} }{2}$.

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