2 I just edited the typesetting and tried to clarify my argument.

$n=2$ is an exception because $S_{(x,2)}$ is periodic in the imaginary part $b$ of $x=a+bi$ with period $\frac{2\pi}{log(2)}$. \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 $a_{2}\frac{2\pi}{log(2)}=a_{3}\frac{2\pi}{log(3)}=a_{n}\frac{2\pi}{log(n)}$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 $a_{2}\frac{2\pi}{log(2)}=a_{3}\frac{2\pi}{log(3)}=a_{n}\frac{2\pi}{log(n)}$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 $a_{2}, ..., a_{n}$ 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.

1

# Complex Zeroes of Stirling functions of the second kind

My motivation to the following question stems from the discussion at http://mathoverflow.net/questions/83999/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 $\frac{2\pi}{log(2)}$. 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 $n=2$?