Suddenly, I encounter this problem: find the sum

$\displaystyle \sum_{n=1}^{N} \sin^2\left(x + \frac{2\pi}{N} n \right)$

Here is my way of thinking,

since

$\displaystyle \sin(a_n) = \frac{1}{2i} (e^{ia_n} - e^{-ia_n} )$

where

$a_n = x + \frac{2\pi}{N} n$

then

$\displaystyle \sin^2(a) = \frac{1}{4} (2- e^{2ia} - e^{-2ia} )$

The sum break down to calculate

$\displaystyle \sum_{n=1}^{N} \sin^2\left(x + \frac{2\pi}{N} n \right) = \frac{N}{2} - \frac{1}{4} \sum_{n=1}^N (e^{2ia} + e^{-2ia})$

The sum is related to n-root of unity as

$\displaystyle \sum_{n=1}^N e^{2ia} = 0$

because

$\displaystyle z^N = 1 \rightarrow z^N-1 = 0 \rightarrow \Sigma_{n=1}^N (z - e^{i 2\pi n / N}) = 0$

break down the product,

$\displaystyle z^N + \sum_{n=1}^N e^{i 2\pi n/N} z^N + .... + \Sigma_{n=1}^N e^{i 2\pi n /N} = 0$

thus,

$\displaystyle \sum_{n=1}^N e^{i 2\pi n/N} = 0$

And since $\sqrt{1} = 1$, thus, $z^{N/2} = 1$ and

$\displaystyle \sum_{n=1}^N e^{i 4\pi n/N} = 0$

The graphical imagination is that, the solutions of n-root of unity are the vectors,  isotropically  distributed from center of a unit circle to its circumference. Thus, the add up of all these vectors will form a N-size polygon, and the result of the sum is zero.

I found it is very interesting to transform one question into another question that can be solved graphically.

This problem is originated from 3-phase AC power that $N = 3$. I wonder, is it the total power be a constant for $N >= 3$, and the answer is yes.

The ultimate reason for that is, when generating AC power, the motor is rotating at constant speed and a constant power is inputting to the system, which can using 1-phase ($N = 2$), 2-phase ($N = 4$), 3-phase ($N = 3$), and so on, to generate the AC power. Thus, as long as the phase distributed equally, the sum of power must be a constant.