Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$. 

$$z^{10}+\frac{1}{z^{10}}=?$$

For the solution:


the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
How do I proceed?
Hint: $z^2+z+1=0$ implies that $0=(z-1)(z^2+z+1)=z^3-1$, so $z^3=1$. Also $z^{10}=z^9\cdot z$ and $z^{-10}=z^2\cdot z^{-12}$.
$z^2+z+1 = 0 \implies   z^2+z = -1 $
$z^2= -z-1 \implies z^3 = -z^2-z = -(z^2+z)= 1$
$z^{10} = z^3\cdot z^3\cdot z^3\cdot z=  z \implies$
$z^{10} + z^{-10} = z + 1/z = \frac{z^2 +1 }z$ but $z^2+1 = -z$ so  this results in $-z/z = -1$