Solving $z^3 =-1$ I need to find the solution of $z^{3} =-1$.
Let$$z=[ r,\theta ]$$
then using de Moivre's theorm$$ z^{3} =\left[ r^{3} ,3\theta \right] $$
write the number $-1$ in modulus argument from,
$$[ 1,( 2n-1) \pi ]$$
the we can write$$\left[ r^{3} ,3\theta \right] =[ 1,( 2n-1) \pi ]$$
Therefore,$$ r^{3} =1 \space \text{and} \space \theta =\frac{( 2n-1) \pi }{3}$$
Then$$z_{n} =1\left( \cos\frac{( 2n-1) \pi }{3} +i\sin\frac{( 2n-1) \pi }{3}\right)$$
\begin{align}
&n=0 \quad z_{0} =\frac{1}{2} -i\frac{\sqrt{3}}{2}\\
&n=1 \quad z_{1} =\frac{1}{2} +i\frac{\sqrt{3}}{2}\\
&n=2 \quad z_{1} =-1\\
\end{align}

$z=re^{i\theta}$
$z^3 = -1$
$r^3 (e^{3 i \theta})= 1\cdot e^{i\pi}$
Hence, $  r=1$ and $\theta= \frac{i(\pi + 2k\pi) }{3}$ for $k=0, 1,2$
$z= e^{\frac{i\pi}{3}} , e^{i\pi}, e^{\frac{5i\pi}{3}}$
