I was exploring the formula:

$$g(s)_{\pm} := \displaystyle \frac{\zeta(\overline{s})}{\zeta(1-s)} \pm \frac{\zeta(s)}{\zeta(1-\overline{s})}$$

and found that for all $\Re(s) \ne \frac12$:

$|g(s)_{+}|$ has pairs of complex zeros that always encapsulate a $\rho$ between them.

$|g(s)_{-}|$ has complex zeros near the known $\rho$'s (call them $\mu$'s).

Similar to the $\rho$'s (if they would exist for $\Re(s) \ne \frac12)$, all $\mu$-roots are equal for $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

The $\mu$'s obviously vanish without a trace when $\Re(s)=\frac12$, however the $\mu$-zeros for $\displaystyle \lim_{\Re(s) \to \frac12}$ clearly and smoothly converge towards the known $\rho$'s.

The $\mu$'s can also be computed by putting $|g(s)_{+}| - 2|\chi(s)|=0$ with $\chi(s)=2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s)$.

It is conjectured (or maybe already proven?), that $\chi(s)$ does not contain any information about the $\rho$'s, however the formula above suggests that for all $\Re(s) \ne \frac12$, there actually is information in $\chi(s)$ about the $\mu$'s. $\chi(s)$ apparently plays a complementary role for $\Re(s)=\frac12$ and $\Re(s) \ne \frac12$.

I realise this is a very broad question, but could this complement in any way help explain that the only way to make $\chi(s)$ fully independent from the $\rho$'s, is when $\Re(s)=\frac12$?


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