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If $1,\alpha _1 ,\alpha _2, .........,\alpha _{2008}$ are $2009$ roots of unity , then the value of $\Sigma ^{2008} _{r=1} r(\alpha _r + \alpha _{2009-r})$

I know sum of all the roots are 0 and product is 1 .

As $x^{2009}-1$

but now how to proceed .

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  • $\begingroup$ $\alpha_r = \alpha_1^r$? And $\alpha_1$ is primitive? $\endgroup$ – Exodd Apr 20 '17 at 12:41
  • $\begingroup$ @Exodd what do you mean by that $\endgroup$ – user123733 Apr 20 '17 at 12:42
  • $\begingroup$ I want to know if $\alpha_r$ are all the 2009 roots of unity, and whether $\alpha_r$ and $\alpha_{2009-r}$ are conjugated $\endgroup$ – Exodd Apr 20 '17 at 12:43
  • $\begingroup$ @Exodd yes they all are roots $\endgroup$ – user123733 Apr 20 '17 at 12:52
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You have that $$\sum_{r=1}^{2008}r(\alpha_r+\alpha_{2009-r})=\sum_{r=1}^{2008}r\alpha_r+\sum_{r=1}^{2008}r\alpha_{2009-r}=\sum_{r=1}^{2008}r\alpha_r+\sum_{r=1}^{2008}(2009-r)\alpha_r$$ Can you take it from here?

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  • $\begingroup$ How you have written $ \sum_{r=1}^{2008}(2009-r)\alpha_r$ $\endgroup$ – user123733 Apr 20 '17 at 12:51
  • $\begingroup$ @user123733 Well instead of starting the sum at $r=1$ to $r=2008$ you start summing from $r=2008$ downwards to $r=1$. $\endgroup$ – kingW3 Apr 20 '17 at 12:55
  • $\begingroup$ So will the answer would be 0 $\endgroup$ – user123733 Apr 20 '17 at 12:57
  • $\begingroup$ Yes,since $2009\cdot 0=0$ $\endgroup$ – kingW3 Apr 20 '17 at 12:58

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