Now, let's express $z$ and $\mathrm{i}$ in polar form. We know that $\mathrm{i} = e^{\frac{\pi}{2}i}$, since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$.
Using De Moivre's theorem, we can find the cube roots of $\mathrm{i}$ as follows:
$z^3 =
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