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$\displaystyle\frac1N\sum_{k=0}^{N-1}e^{\frac{i2\pi\mu k}N}=\begin{cases}1,&k\mid\mu\\0,&k\nmid\mu\end{cases}$

where $\mu=0,\pm1,\pm2,\dots$ and $N>0$.

I hope for the procedure in detail.

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Do you know how to evaluate geometric series? – user27126 Apr 27 '13 at 5:21
sorry,I don't know. I really need your help. – park ning Apr 27 '13 at 5:22
Its not written as a series... – Integral Apr 27 '13 at 5:24
@Integral, it is a geometric series. – user27126 Apr 27 '13 at 5:28
It should be $N|\mu$ – lab bhattacharjee Apr 27 '13 at 5:28
up vote 2 down vote accepted

If $N|\mu, N=r\mu$ (say),$e^{\frac{2k\pi i}N}=e^{2k\pi r i}=1$ for all integer $k$

If $N\not\mid \mu, \sum_{k=0}^{N-1}e^{\frac{i2\pi\mu k}N}$ $=\frac{\left(e^{\frac{i2\pi\mu k}N}\right)^N-1}{e^{\frac{i2\pi\mu k}N}-1}$ $=\frac{e^{i2\pi\mu k}-1}{e^{\frac{i2\pi\mu k}N}-1}=0$ as $e^{\frac{i2\pi\mu k}N}-1\ne0$ as $N\not\mid \mu$

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Hint: In the complex plane, these points are on the unit circle, so if they are evenly spaced, their geometrical average indeed zero. Think about what "evenly spaced" means in terms of $k$ and $\mu$.

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Use geometric series and the properties of complex exponential, you will reduce this expression to some fraction, then try with a divisor of $k$ and another kind of number. This proof is really easy by that way. Fourier

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