Can someone tell me how to prove the folowing equalty : $$\prod_{k=0}^{n-1}e^{\frac{2\pi i k}{n}}=(-1)^{n-1}\;\;\; n\in\mathbb{N}^*.$$
Thanks in advance.
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4Note that $\prod_0^{n-1} e^{ak}=e^{a(0+1+2+\cdots +(n-1))}$. – André Nicolas Jun 18 '14 at 16:40
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@AndréNicolas, Thanks. – Jun 18 '14 at 16:42
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1You are welcome. If you can comfortably finish from there, you may want to give an answer yourself. – André Nicolas Jun 18 '14 at 16:43
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Take the logarithm of the lhs. The resulting expression is just an arithmetic progression. Compute it and go back to the exponential. – Claude Leibovici Jun 18 '14 at 16:46
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I will answer my question
We have : $$\prod_{k=0}^{n-1}e^{\frac{2\pi i k}{n}}=e^{\frac{2\pi i }{n}\frac{n(n-1)}{2}}=e^{\pi i (n-1)}=\cos((n-1)\pi)=(-1)^{n-1}.$$
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Consider the polynomial $x^n-1$. Vieta's formula says that $-1=(-1)^n z_1\cdots z_n$ where the $z_i$ are the roots of the polynomial. What are these roots?
Pedro
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