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If $w = e^{i\frac{2\pi}5} $, then $1 + w + w^{2} + w^{3} + 5w^{4} + 4w^{5} + 4w^{6} + 4w^{7} + 4w^{8} + 5w^{9}$ =?

I substituted $w$ into the expression and combined similar terms. I then tried to see which terns had real or imaginary parts that would cancel. That didn't work out. There's an annoying trick in this problem.

Jyrki Lahtonen
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animalcroc
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1 Answers1

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The trick is that $w^5 = 1$ and $w \neq 1$ implies that

$$1+w+w^2+w^3+w^4 = 0.$$

Use that twice.

Daniel Fischer
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    Oy, I read $e^{2\pi i/5}$. – Daniel Fischer Oct 19 '14 at 20:04
  • I'm fairly sure that this question comes from a practice math subject GRE, where $w$ is indeed $e^{2\pi i/5}$. See #43 of Form GR0568, the most recent practice test from 2008. – Alex Wertheim Oct 19 '14 at 20:13
  • you're right. i just checked, but there is no explanation of the solution. – animalcroc Oct 19 '14 at 21:59
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    @animalcroc Recall that for geometric sums you have $$\sum_{k=0}^{n-1} q^k = \frac{1-q^n}{1-q}$$ if $q \neq 1$. Using that with $q = w$ and $n = 5$ gives you $1+w+w^2w^3+w^4 = \frac{1-w^5}{1-w} = 0$. Write $5w^4= w^4 + 4w^4$ and put the first into the first geometric sum, and the second in the next geometric sum $4(w^4+w^5+w^6+w^7+w^8)$. – Daniel Fischer Oct 19 '14 at 22:05
  • @DanielFischer. how is does it follow that equation is 0? – animalcroc Oct 19 '14 at 22:10
  • wow, what a nasty problem. thanks – animalcroc Oct 19 '14 at 22:12