$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$ Let $z=e^{i\frac{2\pi}{7}}$
Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$
How do I solve from here?

Since $z^7=1$ we have that it's equal to
$$z^2+z^5+z^6+z+z^3+z^4$$
$$=z^6+z^5+z^4+z^3+z^2+z$$ because $z^{7k+1}=z, z^{7k+2}=z^2,...$. This is equal to $-1$ since $z^7-1=(z-1)(z^6+..+1)=0$ and thus $z^6+..+1=0$ noting that $z -1\neq 0$.
Also we can solve using sum of geometric progression.
$\frac{a(r^n-1)}{r-1}=\frac{z(z^6-1)}{z-1}=\frac{z^7-z}{z-1}=\frac{1-z}{z-1}=-1$
