<s>[INST] Solutions of $z^6 + 1 = 0$ Solve:
$$z^6 + 1 = 0$$
That lie in the top region of the plane.
We know that:
$$(z^2 + 1)(z^4 - z^2 + 1) = 0$$
$$z = -i, i$$ 
We need to solve:
$$((z^2)^2 - (z)^2 + 1) = 0$$
$$z = \frac{1 \pm \sqrt{-3}}{2}$$
But this is incorrect. How to do this then?
 [/INST] In order to solve $\sqrt[2n]{-1}$:


*

*Draw the unit circle

*Draw the first solution, which is obviously $0+1i=\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})i$

*Repeat $2n-1$ times: find the next solution by rotating the previous solution $\frac{\pi}{n}$ radians


For example, $\sqrt[6]{-1}$:


*

*$\cos(\frac{ 3\pi}{6})+\sin(\frac{ 3\pi}{6})i$

*$\cos(\frac{ 5\pi}{6})+\sin(\frac{ 5\pi}{6})i$

*$\cos(\frac{ 7\pi}{6})+\sin(\frac{ 7\pi}{6})i$

*$\cos(\frac{ 9\pi}{6})+\sin(\frac{ 9\pi}{6})i$

*$\cos(\frac{11\pi}{6})+\sin(\frac{11\pi}{6})i$

*$\cos(\frac{13\pi}{6})+\sin(\frac{13\pi}{6})i$



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