summing the powers of a complex number Let $z=e^{\frac{2\pi i}{5}}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=?$
I am kind of confused since by drawing a graph, $1+z+z^2+z^3+z^4$ should be zero, but using computational softwares the result is different, and hence I do not know how to solve this problem.
Thanks for helping!

We have $$1+z+z^2+z^3+z^4+5z^4+4z^5+4z^6+4z^7+4z^8+4z^8+5z^9=$$
$$=(1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9.$$
Using the fact that the $\left(e^{2\pi i/5}\right)^5=1$, we see that $z=e^{2\pi i/5}$ is a root of $z^5-1=(z-1)(1+z+z^2+z^3+z^4)$, in particular, it must be a root of $1+z+z^2+z^3+z^4$, so that we are left with $$5z^9=5e^{(2\pi i/5)\cdot 9}=5e^{18\pi i/5}=5e^{8\pi i/5},$$where we have used the fact that $e^{2\pi i}=1$.
