The Zeta-function can be written as the following infinite Hadamard product of its non-trivial zeroes:
$\zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$
this also implies that:
$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{\rho} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$
Take the reflection formula:
$\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$
and substitute the Hadamard products for $\zeta(s)$ and $\zeta(1-s)$. The result is that:
$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$
that can be rewritten as:
$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$
This equation has zeros for $\rho = s$ as long as $2\rho-1 \ne 0$.
The ρ's could obviously lie anywhere in the already proven strip between $0<\Re(\rho)<1$ and I just take them as 'givens' wherever they might be located.
I started to experiment with solving s for different numbers of terms as follows:
$x=\frac13$ (e.g.)
$\prod_{n=1}^y \left(\frac{(x + ni) -s}{(x + ni) + s -1} \right) = 1$
Whatever value I pick for x between 0 and 1, the solution is always a complex number (ignoring $s=\frac12$, that is always a solution). However the exception occurs when $x = \frac12$ that always seems to only produce real numbers as solution(s).
Does anybody see why that is (or must be) the case?
Best Answer
Using the notation $s=u+1/2$ your conjecture can be reformulated and generalized as follows.
Proposition. Let $v_1,v_2,\dots,v_N$ be arbitrary positive numbers, then all solutions of the equation $$ \prod_{n=1}^N \frac{v_ni-u}{v_ni+u} = 1 $$ are real.
Proof. The degree of the polynomial $\prod_{n=1}^N(v_ni-u)-\prod_{n=1}^N(v_ni+u)$ is $N$ or $N-1$ depending on whether $N$ is odd or even (the polynomial is always odd). Therefore it suffices to show that there are the same number of real solutions to the displayed equation. As $u$ grows from $-\infty$ to $\infty$, each fraction under the product traverses the unit circle continuously in the positive direction, starting from and arriving back to $-1$. Using ideas similar to how one proves that the fundamental group of the unit circle is $\mathbb{Z}$, we see that the product traverses the unit circle $N$ times in the positive direction, starting from and arriving back to $(-1)^N$. In particular, the product passes $1$ exactly $N$ or $N-1$ times depending on whether $N$ is odd or even. QED