If $\omega = \cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$
show that $1+\omega^h + {\omega}^{2h}+ \ldots + \omega^{(n-1)h}=0$ if $h$ is not multiple of $n$
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2Treat that as a geometric sum – peterwhy Jan 10 '14 at 16:32
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Another way: http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Jan 10 '14 at 16:42
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Use Euler Formula : $\displaystyle e^{ix}=\cos x+i\sin x$
and summation formula of Geometric Series to get $$\frac{1-w^{nh}}{1-w^h}$$
as $h$ is not multiple of $\displaystyle n, w^h\ne1$
and $w^{nh}=(w^n)^h=\left(\left(e^{\frac{2\pi i}n}\right)^n\right)^h=\left(e^{2\pi i}\right)^h=1^h$
lab bhattacharjee
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Related : De Moivre's formula :http://mathworld.wolfram.com/deMoivresIdentity.html, http://en.wikipedia.org/wiki/De_Moivre's_formula – lab bhattacharjee Jan 10 '14 at 16:40