Question

What are the cube roots of -1+i sqrt 3

Answer 1

\[z^3=-1+i\sqrt3=(r(\cos\theta+i\sin\theta))^3\]
You have that
\[r=\sqrt{(-1)^2+(\sqrt3)^2}=2\]
and
\[
\begin{cases}2\cos\theta=-1\\
2\sin\theta=\sqrt3\end{cases}~~\Rightarrow~~\theta=\frac{2\pi}{3}\]
The roots will have the form
\[z=(-1+\sqrt3)^{1/3}=2^{1/3}\left(\cos\frac{\theta+2k\pi}{3}+i\sin\frac{\theta+2k\pi}{3}\right)\]
for \(k=0,1,2\).