Complex Zeroes of Stirling functions of the second kind My motivation to the following question stems from the discussion at Zeros of "exponential" function about the real zeroes of Stirling numbers of the second kind, I am curious in exploring the complex zeroes of Stirling functions of the second kind.
Define: $S_{(x,n)}=\frac{1}{n!}\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{x}$ for integer $n$ and complex $x$.
I conjecture that for $n>2$, $S_{(x,n)}$ has exactly $n-1$ complex zeroes. I realize this is a much stronger claim than the fact that $S_{(x,n)}$ has $n-1$ real zeroes. However, I have noticed after examining a large amount of mathematical data that for $x>n$, both the real parts and the imaginary parts of $S_{(x,n)}$ seem to alternate for increasing integer $n$ when $x$ is a complex number. 
When $x$ is a real number, the imaginary part of $S_{(x,n)}$ is zero and we can say that the imaginary part alternates as well for real $x$ from $+0=0$ to $-0=0$. My question becomes the following. If we consider $S_{(x,n)}$ for complex $x$ as a linear combination of integer exponential functions that are defined for complex $x$, does it follow that for $n>2$, $S_{(x,n)}$ has exactly $n-1$ complex zeroes?
$n=2$ is an exception because $S_{(x,2)}$ is periodic in the imaginary part $b$ of $x=a+bi$ with period $\frac{2\pi}{\log{3}}$. This is an exception because the equation $S_{(x,2)}=0$ is equivalent to the equation $2^{x}-2=0$ which only has one exponential term that is non-constant. Since $a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$ presumably does not have a solution for $a_{2}, ..., a_{n}$ all integers except for when $n=2$ it is unlikely that there are any other complex zeroes of $S_{(x,n)}$ in the case that $n\ne 2$ other than $x=1$, ..., $n-1$.
Can someone help me think of a way to show that $a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$ does not have a solution where $a_{2}, ..., a_{n}$ are all integers except for when $n=2$? 
Furthermore, I want to clarify the fact that I am only considering whether $S_{(x,n)}$ is periodic in the imaginary part of $x=a+bi$, because for any complex zero of $S_{(x,n)}$, x=a+bi, $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$.
We note that in each equation $a$ and $b$ are real. Therefore, as Elkies already argued, the maximum number of real zeroes of an equation $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ or $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$ with $n-1$ sign changes is $n-1$. In each case, there are at maximum $n-1$ values of $a$ that satisfy the equation. For each value of $a$, it can be observed that there is at most $1$ value of $b$ that satisfies both equations (this is equivalent to the fact that $a_{2}\frac{2\pi}{\log{2}}=a_{3}\frac{2\pi}{\log{3}}=a_{n}\frac{2\pi}{\log{n}}$ does not have a solution where $a_{2}, ..., a_{n}$ are all integers except for when $n=2$).
Thank you for your help in advance. I hope this makes my line of reasoning more clear.
 A: Your function has infinitely many complex zeros.
Indeed, it is of the form
$$\sum_{k=1}^n a_k e^{\lambda_k z},$$
where $a_k$ are constants, and $\lambda_k=\log k$ are all distinct.
A function of this form always has infinitely many zeros, unless
$n=1$.
Proof. This is an entire function of order $1$, normal type. So, if it
has finitely many zeros, then by Hadamard's factorization theorem,
it must be of the form $P(z)e^{cz}$, where $P$ is a polynomial,
and $c$ is a constant.
Thus we have
$$\sum_{k=1}^n a_k e^{(\lambda_k-c)z}\equiv P(z).$$
By differentiating this sufficiently many times to kill $P$, we will
obtain that exponentials with distinct exponents are linearly
dependent, while everyone knows that this is not so.
(For the proof, consider the Wronski determinant, after division by
an exponential it will become a Wandermonde determinant) and we know
that Wandermonde determinant of distinct numbers is never $0$).
A: $This \ is \ not \ an \ answer$.
This is merely too long to be a comment to Alexandre Eremenko's response.
Let me rephrase my comment. I managed to show before that by using Vandermonde matrices that the vectors $(1^{x}, ..., n^{x})^{T}, (1^{1}, ..., n^{1})^{T}, (1^{2}, ..., n^{2})^{T}, ..., (1^{n-1}, .., n^{n-1})$ are linearly dependent precisely when $S_{(x,n)}=0$. Therefore, $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$ for at most $n-1$ values of $a$ for a fixed $b$. This can be seen because by acknowledging the fact that $(1^{x}, ..., n^{x})^{T}, (1^{1}, ..., n^{1})^{T}, (1^{2}, ..., n^{2})^{T}, ..., (1^{n-1}, .., n^{n-1})^{T}$ are linearly dependent for all real $x$ $x\ne 1, ..., n-1$, then we can conclude that $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\cos{(b\log{k})}=0$ and $\sum\limits_{k=1}^{n}{n\choose k}(-1)^{n-k}k^{a}\sin{(b\log{k})}=0$ has at most $n-1$ zeroes because it can be written $({n\choose 1}(-1)^{n-1}\cos{(b\log{1})}, ...,{n\choose n}(-1)^{n-n}\cos{(b\log{n})})^{T} \cdot (1^{a}, ..., n^{a})^{T}$ and $a$ is a real number.
Is this following logic correct? This is a more plausible fact (which I hope is true).
