<s>[INST] Gre Question Complex Number (plug and chug) This seems like it should be easy, but I can't seem to simplify it: If $z=e^{i\frac{2\pi}{5}}$, then what is $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$. The choices are $0, 4e^{i\frac{3\pi}{5}}, 5e^{i\frac{4\pi}{5}}, -4e^{i\frac{-2\pi}{5}}, -5e^{i\frac{3\pi}{5}},$ with the answer being $-5e^{i\frac{3\pi}{5}}.$ I can plug in the given $z$ into the equation and get $5+10e^{-i\frac{2\pi}{5}}+5e^{i\frac{2\pi}{5}}+5e^{i\frac{-4\pi}{5}}+5e^{i\frac{4\pi}{5}}$, but have been unsuccessful in simplifying it so far.
 [/INST] Note that $z^5=1$, as $z$ is a fifth root of unity, so the expression simplifies to
$$
\begin{align}
1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4 &=5+5z+5z^2+5z^3+10z^4\\
&=5(1+z+z^2+z^3+z^4)+5z^4
\end{align}
$$
However, either by using the formula for the geometric series, or the fact that $1+X+X^2+X^3+X^4$ is the fifth cyclotomic polynomial, it follows that $1+z+z^2+z^3+z^4=0$. So the final answer is
$$
5z^4=5\exp(i8\pi/5)=-5\exp(i3\pi/5).
$$
</s>