# The Roots of Unity and the diagonals of the n-gon inscribed in the unit circle

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.

• You cannot simply remove the absolute sign without checking that you have a real number (which you don't) – Calvin Lin Jan 26 '13 at 23:06
• A good approach to this problem is to use vectors instead. Do you know the dot product? – Calvin Lin Jan 26 '13 at 23:08
• Yes. How could I use it? – Ryuky Jan 26 '13 at 23:08
• Read Worked Example 3 and then do Test Yourself 3. – Calvin Lin Jan 26 '13 at 23:10
• 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? – Ryuky Jan 26 '13 at 23:15

## 2 Answers

Hint: Use the correct expansion $$|1-\omega^k|^2=(1-\omega^k)(1-\overline{\omega}^k)=(1-\omega^k)(1-\omega^{-k}).$$ IOW $|z|^2=z\cdot\overline{z}$.

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.