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}

You can find the complex roots of unity as below. The equation you have solved is not the one that will give you complex roots of unity.
$$z^3=1 \implies z^3-1=0$$
Factor the LHS (using any of a variety of methods):
$$(z-1)(z^2+z+1)=0$$
Use the zero-product property:
$$z-1=0\implies z=1$$
$$z^2+z+1=0\implies z=\frac{-1\pm i\sqrt{3}}{2}$$
(where we used the quadratic formula above)
