News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

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 :

Choice 5 is the answer.

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}.


Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.