# If $z=e^\frac{2\pi i}{5}$, it is true that $1+z+z^{2}+z^{3}=e^\frac{3\pi i}{5}$?

If so, how can one show this? $1+z+z^{2}+z^{3}=\frac{z^4-1}{z-1}=\frac{\overline{z-1}}{z-1}$ is as far as I can get.

There's a nice fact that the sum of the roots of unity are $0$. In this case, where $z$ is the $5$th root of unity, you have that $1 + z + z^2 + z^3 + z^4 = 0$, so you find that $$1 + z + z^2 + z^3 = -z^4 = -e^{8\pi i/5} = -e^{\pi i}\cdot e^{3\pi i/5} = e^{3\pi i/5}$$ as desired!
$z$ is a $5$th root of unity ($z^5 = e^{10\pi i/5} = e^{2\pi i} = 1$). It's well-known that for an $n$th root of unity $r_n$ that: $$\sum_{i = 0}^{n-1} r_n^i = 0$$ So, we have that: $$1 + z + z^2 + z^3 + z^4 = 0$$ So, we have that: $$1 + z + z^2 + z^3 = -z^4 = -e^{8\pi i / 5} = e^{\pi i}e^{8\pi i/5} = e^{13\pi i/5} = e^{3\pi i/5 + 2\pi i} = e^{3\pi i/5}$$