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I want to prove that the sum of the squares of the diagonals of a regular $n$-gon inscribed in the unit circle is equal to $2n$. So what I've done is I considered the $n$th roots of unity and said that the diagonals are given by $$|1-\omega^k|$$ where $k=1,2,3,...,(n-1)$.

Then the sum of the squares is given by $$\sum_1^{n-1}|1-\omega^k|^2=\sum_1^{n-1}(1-2\omega^k +\omega^{2k})=n-1 -2\sum_1^{n-1}\omega^k + \sum_1^{n-1}\omega^{2k}.$$

The first sum is -1, right? But what about the second? Since the roots are repeating shouldn't the sum be equal to the first sum? But if so then the result is not $2n$ as I have to show...

Any help will be appreciated.

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  • $\begingroup$ You cannot simply remove the absolute sign without checking that you have a real number (which you don't) $\endgroup$ – Calvin Lin Jan 26 '13 at 23:06
  • $\begingroup$ A good approach to this problem is to use vectors instead. Do you know the dot product? $\endgroup$ – Calvin Lin Jan 26 '13 at 23:08
  • $\begingroup$ Yes. How could I use it? $\endgroup$ – Ryuky Jan 26 '13 at 23:08
  • $\begingroup$ Read Worked Example 3 and then do Test Yourself 3. $\endgroup$ – Calvin Lin Jan 26 '13 at 23:10
  • $\begingroup$ I understand the proof. However the question I have to answer requires that I use complex numbers and the roots of unity. The absolute sign is there because it is the distance from 1 to the first, second, ... (n-1) root of unity. Doesn't the square take care of it anyway? $\endgroup$ – Ryuky Jan 26 '13 at 23:15
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Hint: Use the correct expansion $$ |1-\omega^k|^2=(1-\omega^k)(1-\overline{\omega}^k)=(1-\omega^k)(1-\omega^{-k}). $$ IOW $|z|^2=z\cdot\overline{z}$.

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Your equation is incorrect, because $ | 1 - \omega|^2 \neq (1-\omega)^2$, as the RHS is not a real number.


$|1 - \omega|^2 = |1 - \omega | | 1 - \overline{\omega} | = 1 - (\omega + \overline{\omega}) + \omega \overline{\omega} $.

Now use the fact that $ x^n -1 = 0$, which tells you the sum of roots of unity and the sum of cross terms of roots of unity.

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