Finding the sum of the real parts squared of the roots of unity

Question

I was solving a diophantine equation when I reached this point and I got stuck finding the value of
$$ \sum_{k=1}^n \cos^2\left(\frac{360^{\circ}}n k\right).$$ I noticed that this sum resembles the sum of the real parts squared of the roots of unity.
After graphing the function and/or plugging in some values you can quickly realize that $$\sum_{k=1}^n \cos^2\left(\frac{360^{\circ}}n k\right) = \frac n2$$ for all natural numbers greater than $2.$

How can we prove this?

https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Oct 03 '20 at 06:24

score 3 · Accepted Answer · answered Oct 03 '20 at 06:27

3

$$\sum_{k=1}^n\cos^2(2\pi k/n)=\frac{1}{2}\sum_{k=1}^n(\cos(4\pi k/n)+1)=\frac{n}{2}$$ since $\sum_{k=1}^n\cos(4\pi k/n)=\mathrm{Re}\sum_{k=1}^ne^{4\pi ki/n}=0$.

answered Oct 03 '20 at 06:27

Chrystomath

10,798

why is this true $ \mathrm{Re}\sum_{k=1}^ne^{4\pi ki/n}=0$ – Ak2399 Oct 03 '20 at 06:37
1

$\sum_{k=0}^{n-1}e^{4\pi ik/n}=1+\omega+\cdots+\omega^{n-1}=\frac{1-\omega^n}{1-\omega}=0$ since $\omega^n=e^{4\pi i n/n}=1$. – Chrystomath Oct 03 '20 at 06:39

Finding the sum of the real parts squared of the roots of unity

1 Answers1