It has been proven that:

1) if $s$ is a non trivial zero $\rho$ of $\zeta(s)$ then so is $1−s$.

2) $\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$

3) $ 0 < \Re(\rho) <1$

From this it follows that when $s \to \rho$:

$\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}| = |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$

It is easy to see that the outcome will be $1$ for all $y$ in $s=\frac12 + y i$.

But if a $\rho$ would lie off this critical line, it also must reside in 'spots' where $\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}|=1$.

On which points off the critical line could this occur? I found a surprisingly small domain (no proof).

The blue line shows the only values where:

$\displaystyle |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$, $s=x + y i$, $ 0 \le x \le 1$.

Note that $y \to 2\pi$ for both $x=0$ and $x=1$. The $y$ rises only a little in the middle.

This doesn't say anything about whether or not off-line $\rho$'s are actually hiding on this curve. There still is an infinite number to check. However, I wondered if anything more is known about this curve?

[IMG]http://img822.imageshack.us/img822/3065/riemanntest.jpg[/IMG]