Take the 2-minute tour ×  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