First, is there a better way than using x+iy and solving the system? I tried letting $z=e^{i3\theta}$ and using the cosine and i*sine way but I don't see how that can equal -1 and i at the same time, either cosine is -1 or sine is 1 not both?
Anyway I've arrived at the real and imaginary parts by plugging in $x+iy$ to get the system
$$x^3-3y^2x=-1$$
$$3yx^2 -y^3=1$$
We can conclude y and x are nonzero but how do we solve?
Best Answer
You can't put $z^3=-1+i=e^{3i\theta}$ because $\lvert -1+i\rvert = \sqrt{2}$ but $\lvert e^{3i\theta}\rvert = 1$, you need to normalize $z$, i.e. $z=re^{i\theta}$, $r\in\mathbb{R}$. Then $r^3 = \sqrt{2}$ and after that, solve for $\theta$.