Find the $n$-th roots of $z^n +2=0$

Question

I'm trying to solve the complex equation $z^n +2=0$, in terms of $e$, so my first approach is using De Moivre's formula:

$z^n +2=0$

$z^n=-2$

$z=-2^{1/n}$,

But I don't know how to express it for put them in the unit circle.

Thanks

Answer 1

$z^n=2e^{i\pi}$

$z=2^{\frac{1}{n}}e^{i\frac{2k\pi+\pi}{n}}$, $k=0,1,\cdots, n-1$.