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.

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