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Let $x_1, \ldots,x_n$ be complex numbers such that for any $k$, $$ \sum_{i=0}^n x_i^k = 0.$$

I'd like to show that this implies $x_1 = x_2 = \cdots = x_n = 0.$

I was suggested to use this strategy. Suppose $|x_1| \geq |x_2| \geq \cdots \geq |x_n|.$ Then $$ \lim_{k\to \infty} \left(\sum_{i=0}^n x_i^k \right)/x_1 = d$$ where $d$ is the total of all numbers with maximal absolute value.

This would then imply that for large enough $k$ the sum from the proposition is not zero.

What I was wondering is, is there a more straightforward way of showing this?

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Consider $f(t) = \sum_k t^k \sum_i x_i^k$. – Qiaochu Yuan May 24 '12 at 23:27
Closely related to…. – lhf May 25 '12 at 2:44
up vote 5 down vote accepted

Consider the polynomial with roots as $x_1,x_2,\ldots,x_n$ i.e. $$p_n(z) = (z-x_1)(z-x_2)(z-x_3)\cdots (z-x_n)$$ and use Newton's identities to conclude that the polynomial is $z^n$.

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