# Zeros of a complex function

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. I think the answer is affirmative.

• Not for $n=0$, no. Jul 4 at 0:55

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 $$c>0$$.