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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.

Later
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inZugzwang
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2 Answers2

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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.

Calvin Lin
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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 $|z|^2=z\cdot\overline{z}$.

Jyrki Lahtonen
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