<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] Hint $\rm\ \ z\ne 1,\,\ z^5 = 1\:\Rightarrow\: (\color{#C00}{1\!+\!z\!+\!z^2\!+\!z^3\!+\!z^4})(1\!+\!4z^4)+5z^9 =\, \color{#C00}0+5z^9 =\, 5/z$
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