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?
 A: Only your last line is incorrect. What you should write is
$$z^2 = \frac{1\pm\sqrt{-3}}{2}$$
A: $$z^6=-1=e^{(2n+1)\pi i}$$ 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 $>0$
$\implies\sin\dfrac{(2n+1)\pi}6>0\implies0<\dfrac{(2n+1)\pi}6<\pi\iff0<2n+1<6\implies-.5< n<2.5$
$\implies n=0,1,2$
A: The simplest way of solving this equation is the method based on DeMoivre's Formula that Lab Bhattacharjee outlined.  
That said, you can make your method work.  You found the roots $z \pm i$ by setting the factor $z^2 + 1$ equal to zero.  As Rasolnikov and 5xum noted, you should have obtained 
$$z^2 = \frac{1 \pm \sqrt{-3}}{2}$$
when you set the factor $z^4 - z^2 + 1$ equal to zero.
Let $z = a + bi$, with $a, b \in \mathbb{R}$.  Then 
\begin{align*}
z^2 & = \frac{1 \pm i\sqrt{3}}{2}\\
(a + bi)^2 & = \frac{1 \pm i\sqrt{3}}{2}\\
a^2 + 2abi - b^2 & = \frac{1 \pm i\sqrt{3}}{2}
\end{align*}
Equating real and imaginary parts yields
\begin{align*}
a^2 - b^2 & = \frac{1}{2}\tag{1}\\
2ab & = \pm\frac{\sqrt{3}}{2}\tag{2}
\end{align*}
Solving equation 2 for $b$ yields
$$b = \frac{\pm\sqrt{3}}{4a}\tag{3}$$
Substituting this expression in equation 1 yields
\begin{align*}
a^2 - \frac{3}{16a^2} & = \frac{1}{2}\\
16a^4 - 3 & = 8a^2\\
16a^4 - 8a^2 & = 3\\
16a^4 - 8a^2 + 1 & = 4 && \text{complete the square}\\
(4a^2 - 1)^2 & = 4\\
4a^2 - 1 & = \pm 2\\
4a^2 & = 3 && \text{since $a \in \mathbb{R}$}\\
a^2 & = \frac{3}{4}\\
a & = \pm \frac{\sqrt{3}}{2}
\end{align*}
Substituting this expression into equation 3 yields the four roots 
$$z = \pm\frac{\sqrt{3}}{2} \pm \frac{1}{2}i$$
of the equation $z^4 - z^2 + 1 = 0$. As you can check, these roots correspond to the values $n = 0, 2, 3, 5$ in the formula Lab provided.   
A: 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, $\sqrt[6]{-1}$:


*

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

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

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

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

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

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



