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I wonder whether $\sum_{k=0}^n \exp(r_k z)$ has a complex zero for any $n\in \mathbb{Z}_n^*,0=r_0<r_1<r_2<\dotsb<r_n$. I think the answer is affirmative.

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    $\begingroup$ Not for $n=0$, no. $\endgroup$ Jul 4 at 0:55

2 Answers 2

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An affirmative answer follows from (9) in this paper by Ritt.

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A simpler proof can be obtained as follows. Proving by contradiction, suppose it has no zeros. Since this is an entire function of order one, it must be $\exp(az+b)$. So we have the identity $$\sum_{k=0}^n e^{r_kz}=e^{az+b},$$ where all $r_k$ are distinct. If $a\not\in\{ r_k\}$, this gives a linear dependence of $n+2$ exponential functions. If $a\in\{ r_k\}$, this gives a linear dependence of $n+1$ exponential functions. But it is well-known and easy to prove (using Wronskian determinant) that exponential functions with distinct exponents are linearly independent.

Remark. A little more work shows that in fact such a sum has infinitely many zeros, moreover the sequence of zeros has non-zero density: the number of zeros in disks $|z|\leq r$ is at least $cr$ for some $c>0$.

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    $\begingroup$ Beautiful proof. $\endgroup$
    – Ke Shi
    Jul 4 at 6:07

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