Solve for z^8-z^4+1 = 0

Question

Current Work

This is what I currently have. I'm not sure how to continue. Can someone show or teach me how it's done?

Edit: Work:

\begin{array}{*{20}c} {z = \frac{{ - 1 \pm \sqrt {1-4} }}{{2}}} \\ \end{array}

\begin{array}{*{20}c} {z = \frac{{ - 1 \pm{i}\sqrt {-3} }}{{2}}} \\ \end{array}

$$z^4 ={r^4(\cos(4θ))(\sin(4θ)}$$

Answer 1

If $z^4=-1$ then $z^8-z^4+1=3\ne 0$, so that's out. Multiplying by $z^4+1$ gives the equivalent result $(z^{4})^3+1=0$, so $-z^4$ is a third root of unity distinct from $1$. Thus $z^4=\exp(\pi i\pm\frac{2\pi i}{3}),\,z=\exp(\frac{\pi i}{4}\pm\frac{\pi i}{6}+n\frac{\pi i}{2})$ with $n\in\{0,\,1,\,2,\,3\}$.

Answer 2

If $a + bi = re^{i\theta}= r\cos \theta + r\sin \theta i$ then $\frac ba = \frac {\sin \theta}{\cos \theta}=\tan \theta$ so $\theta = \arctan \frac ba$

And you have $z^8 -z^4 +1 = 0$ so $z^4 = \frac {1 \pm \sqrt{- 3}}{2}$

So $z^4 =\frac {1\pm i\sqrt 3}{2} = \sqrt {1 + 3} e^{ i\arctan \pm\frac {\sqrt 3}1}=2e^{\pm i\frac {\pi}3}$

So $z = \sqrt[4]{2}e^{(\pm i\frac {\pi}{12} + k\frac {\pi i}4)}$

Answer 3

If you let $z^4=y$ you have $y^2-y+1=0$ which you can feed to the quadratic formula to get $$y=\frac 12(1\pm i\sqrt 3)$$ Now find the fourth root of each of these $y$s and you are there.