# 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.

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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 $$5z^4=5\exp(i8\pi/5)=-5\exp(i3\pi/5).$$
 Thank you, that's perfect. – JLA Aug 7 '12 at 8:07 Glad to help. ${}$ – yunone Aug 7 '12 at 8:08
Hint $\rm\ \ z\ne 1,\,\ z^5 = 1\:\Rightarrow\: (\color{#C00}{1\!+\!z\!+\!z^2\!+\!z^3\!+\!z^4})(1\!+\!4z^4)+5z^9 =\, \color{#C00}0+5z^9 =\, 5/z$