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!

According to wolframalpha you are right about $1+z+z^2+z^3+z^4 = 0$
For the sum :
$1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$
$(1+z+z^2+z^3+z^4)+4z^4+4z^5+4z^6+4z^7+4z^8+5z^9$
$0+4z^4(1+z+z^2+z^3+z^4)+5z^5*z^4$
$4z^4(0)+5*1*z^4$
$5*z^4$
