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## Tuesday, 30 August 2011

### Math GRE - #29

If $z=e^{2\pi i/5},$ then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$
1. $0$
2. $4e^{3\pi i/5}$
3. $5e^{4\pi i/5}$
4. $-4e^{2\pi i/5}$
5. $-5e^{3\pi i/5}$

Solution :

Recall that the sum of all the $n-th$ roots of unity for $n>1$ ($n=5$ in this case) is 0. For this particular question, this means that $1+z+z^2+z^3+z^4 = 0.$ This motivates us to write the sum in the question as: $\begin{eqnarray*} (1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9 & = & 0+4z^4\cdot0+5z^9 \\ & = & 5z^9. \end{eqnarray*}$ Now we can easily evaluate the sum: $\begin{eqnarray*} 5z^9=5\left(e^{2\pi i/5}\right)^9 & = & 5\left(e^{2\pi i/5}\right)^4\cdot\left(e^{2\pi i/5}\right)^5 \\ & = & 5\left(e^{2\pi i/5}\right)^4 \\ & = &5e^{8\pi i/5}=-5e^{3\pi i/5}. \end{eqnarray*}$
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