<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] $$z^6=-1=e^{(2n+1)\pi i}$$ where $n$ is any integer
$$\implies z=e^{\dfrac{(2n+1)\pi i}6}=\cos\dfrac{(2n+1)\pi}6+i\sin\dfrac{(2n+1)\pi}6$$ where $0\le n\le 5$
Top region of the plane, $\implies$ the ordinate has to be $>0$
$\implies\sin\dfrac{(2n+1)\pi}6>0\implies0<\dfrac{(2n+1)\pi}6<\pi\iff0<2n+1<6\implies-.5< n<2.5$
$\implies n=0,1,2$
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