This is a follow-up question to this :
Why does $\sum a_i \exp(b_i)$ always have root?
question in link :
Let $z$ be complex. Let $a_i,b_i$ be polynomials of $z$ with real coefficients. Also the $a_i$ are non-zero and the non-constant parts of the polynomials $b_i$ are distinct.
Let $j > 1$.
$$f(z) = \sum_{i=1}^j a_i \exp(b_i)$$
It seems there always exists a complex value $s$ such that
$$ f(s) = 0$$
Is this true? If so, why? How to prove this? If false, what are the simplest counter-examples?
This was the answer :
This follows from the theory of entire functions of finite order in complex analysis. Specifically, we have:
Proposition: Suppose $f(z) = \sum_{i=1}^j a_i \exp b_i$ for some polynomials $a_i,b_i$ (which may have complex coefficients, though the question specifies real polynomials). Then if $f\,$ has no complex zeros then there exists a polynomial $P$ such that $f = \exp P$.
Proof: let $d = \max_i\max(\deg a_i,\deg b_i)$. If $d \leq 0$ then $f$ is constant and we may choose for $P$ a constant polynomial. Else there exists a constant $A$ such that $\left|\,f(z)\right| \leq \exp(A\left|z\right|^d)$ for all complex $z$. This makes $f$ an entire function of order at most $d$. If $f$ has no zeros then $f = e^g$ for some analytic function $g$, and it follows that $g$ is a polynomial (by a special case of the Hadamard product for an entire function of finite order). $\Box$
Moreover, once we put the expansion $f(z) = \sum_{i=1}^j a_i \exp b_i$ in normal form by assuming that each $b_i$ vanishes at zero (else subtract $b_i(0)$ from $b_i$ and multiply $a_i$ by $e^{b_i(0)}$), then at least one of the $b_i$ is $P-P(0)$, and we can cancel and combine terms to identify $f$ with $\exp P$. The proof (by considering behavior for large $|z|$) is somewhat tedious, though much easier in the real case [hint: start by writing $f(z) \, / \exp P(z)$ as $\sum_{i=1}^j a_i \exp (b_i-P)$]. In particular, if $j>1$ and no two $b_i$ differ by a constant then $f$ cannot equal $\exp P$ and thus must have complex zeros.
Now the thing is this answer is correct because the function is entire.
But an infinite sum of exponentials is not always entire.
In particular zeta functions or more general Dirichlet functions.
So consider $ f(z) = \sum a_n n^{-z} $ with a pole at $s=1$ and the $a_n$ being real.
Now it seems there are always zero's. In particular nonreal ones. And they are nontrivially distributed. In fact there are infinitely many nonreal nontrivial zero's for all such $f(z)$.
But why is that ?
Or is that not always so ?
A less general conjecture is that if $f(z)(z-1)$ is entire ( no poles, natural boundaries or singularities) then $f(z)(z-1)$ has infinitely many nonreal nontrivial zero's.
I think the less general conjecture implies to full ( because singularities take many values ).
How to prove that
if $f(z)(z-1)$ is entire then $f(z)(z-1)$ has infinitely many nonreal nontrivial zero's.
I heard about " the universality of the zeta function " but I did not fully understand. Not sure if that is related.
I read a book on zeta functions but found no answer.
Informally I would say because the function reaches local max and local min so the derivative has zero's.
But every dirichlet series is an integral of a dirichlet series.
Not sure if that is even a good argument.
Analytic continuation makes things harder.