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I need to sum$$\sum_{k=1}^{n-1} |1- e^{{2\pi ik}\over {n}}| $$ which finally reduces to

$$\sum_{k=1}^{n-1} 2\sin\ {{\pi k} \over {n}}.$$

But I'm stuck here.The final answer is supposed to be $n$ .

Davide Giraudo
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donvoldy666
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  • Use http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Apr 18 '13 at 10:36
  • I did, but got some weird answers. – donvoldy666 Apr 18 '13 at 10:39
  • What are the weird answers? – lab bhattacharjee Apr 18 '13 at 10:43
  • 2.cot(pi/2n), yeah just one answer,no 'answers' . – donvoldy666 Apr 18 '13 at 10:47
  • I reached at the same value. Why is it weird? – lab bhattacharjee Apr 18 '13 at 10:50
  • Because the answer is supposed to be 'n'. – donvoldy666 Apr 18 '13 at 10:51
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    Well, unless I made a mistake, for $,n=3,$ we have

    $$\left|1-e^{\frac{2\pi i}{3}}\right|+\left|1-e^{\frac{4\pi i}{3}}\right|=2\left|\frac{3}{2}-\frac{\sqrt 3}{2}i\right|=\sqrt{9+3}=2\sqrt 3\neq 3$$

    so your claim is not true...Check this, either you or I are wrong.

     – DonAntonio Apr 18 '13 at 10:57
  • To be honest the original question asked me to prove that the sum of the distances of the points $e^{{2\pi ik}\over {n}}$ k=1,2,...(n-1) (on a unit circle) from 1 was equal to n . – donvoldy666 Apr 18 '13 at 11:04
  • Oh, I see. Well, then there's no doubt: the claim's false since, say in the example I gave above, the three roots of unit form an equilateral triangle, so the distance you're looking for is just the length of any of its sides, which is $,\sqrt 3,$ ...the claim comes closer to be true if instead you want the sum of their distances from the origin, but then it is trivially true that the sum of all this distances (up and including $,1,$) is $,n,$ ... – DonAntonio Apr 18 '13 at 11:19

2 Answers2

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Hint

$$\sum_{k=1}^{n-1} 2\sin\ {{\pi k} \over {n}}=2\mathrm{Im}\left(\sum_{k=1}^{n-1}e^{ik\pi/n}\right)$$ and we can sum the terms of geometric sequence.

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I think the question should be the value of the following sum:

$$\sum_{k=0}^{n-1}\left (1-e^{\frac{2\pi ik}{n}} \right ) = \sum_{k=0}^{n-1}1 - \sum_{k=0}^{n-1}e^{\frac{2\pi ik}{n}} = n - \frac{1-e^{2\pi i}}{1-e^{\frac{2\pi i}{n}}} = n$$

The last equality follows because $e^{2\pi i}=1$.

Matt L.
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