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Finally I got two equations:

\begin{align} z^3&=-\sqrt{3}+i \tag1 \label1 \\ z^3&=-\sqrt{3}-i \tag2 \label2. \end{align}

So there are six roots. It is easy to get two of them: $z_1=(-\sqrt{3}+i)^{1/3}$ for \eqref{1} and $z_4=(-\sqrt{3}-i)^{1/3}$ for \eqref{2}.

However, I am not sure how to apply the rotatation to get the rest of the roots? According to answer the roots are: \begin{alignat}{2} z_2&=\frac {1} {2}i(-\sqrt{3}+i))^{4/3} \quad& z_3&=(-1)^{2/3}(-\sqrt{3}+i)^{1/3} \\ z_5&=-(\sqrt{3}+i)^{1/3} & z_6&=-\frac {1} {2}i(-\sqrt{3}-i)^{4/3}. \end{alignat} Any hints are welcome.

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  • $\begingroup$ Are you familiar with DeMoivre's theorem? $\endgroup$ – Hyperion Feb 23 '19 at 8:12
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    $\begingroup$ @Hyperion Yes, thank you. I tried it and got the roots which have cos and sin functions with 5pi/6 angle. However, it is far from the answers $\endgroup$ – Mikhail Gaichenkov Feb 23 '19 at 8:18
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We can apply DeMoivre's theorem to solve for $z^3 = -\sqrt{3} + i$. Rewriting in polar form, we find that $-\sqrt{3} + i = 2(\cos(5\pi/6) + i\sin(5\pi/6))$.

Applying DeMoivre's theorem to get our first cubic root, we find that $$z_1 = 2^{1/3}(\cos(5\pi/18) + i\sin(5\pi/18))$$.

To get our second $z$ value, we need to rotate our answer around the pole by $2\pi/3$, then apply this rotation once more to get our third $z$. Rotations correspond to "adding angles" in polar form, and thus, our other two roots are

$$z_2 = 2^{1/3}(\cos(17\pi/18) + i\sin(17\pi/18))$$ $$z_3 = 2^{1/3}(\cos(29\pi/18 ) + i\sin(29\pi/18))$$

You can try to apply this principle yourself to get the last $3$ roots from your other equation.

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