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It is easy to see that the function:

$$\zeta(s) \pm \dfrac{1}{\zeta(1-s)}$$

has a pole at each non-trivial zero $s=\rho_n$.

However, after some experiments with this function, I would like to conjecture that:

$$|\zeta(s) - \dfrac{1}{\zeta(1-s)}|$$

only has zeros in the critical strip on the line $\Re(s)=\frac12$, but also that:

$$|\zeta(s) + \dfrac{1}{\zeta(1-s)}|-2$$

always has at least a zero in the critical strip for each $\Re(s) \ne \frac12$ (i.e. all zeros lie off the critical line).

Since both conjectures complement each other, I guess they must be connected in some way.

Using the reflection formula with $\chi(s)=2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \phantom. \Gamma(1-s)$, I rewrote the functions as:

$$\left| \dfrac{\zeta(1-s)^2 \chi(s) - 1}{\zeta(1-s)\chi(s)} \right| =0 \text { and } \left| \dfrac{\zeta(1-s)^2 \chi(s) + 1}{\zeta(1-s)\chi(s)}\right| = 2$$

however this didn't help much solving e.g. the first conjecture that implies $\zeta(1-s)^2 \chi(s) = 1$ only when $\Re(s)=\frac12$ in the critical strip.

Grateful for any steers/hints on how I could best approach this problem.

Thanks!

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  • $\begingroup$ Agno, your second conjecture is much harder :-). Couldn't find small counterexamples. Try plotting the function for large ranges and look for negative values (if they would exist). $\endgroup$
    – joro
    Commented Dec 1, 2012 at 8:01

1 Answer 1

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I suppose these are at least 58 counterexample examples for the conjecture about the zeros of $|\zeta(s) - \dfrac{1}{\zeta(1-s)}|$ in the critical strip with $0 < \Im(s) < 306$.

The first few are:

 0.83937756715930464851 + 31.699222084299394447 i
 0.16062243284069535149 + 31.699222084299394447 i
 0.11131948892305866087 + 88.155391355655174901 i
 0.88868051107694133913 + 88.155391355655174901 i
 0.89823546410519697899 + 123.67527158830200698 i
 0.24612771379601443481 + 122.093560045649621 i
 0.10176453589480302101 + 123.67527158830200698 i
 0.75387228620398556519 + 122.093560045649621 i
 0.26796662801293497366 + 134.16024368224812469 i
 0.73203337198706502634 + 134.16024368224812469 i
 0.86157401152020623657 + 140.44980499028326033 i
 0.86157401152020623376 + 140.44980499028326032 i
 0.13842598847979376624 + 140.44980499028326032 i
 0.89589742682348946277 + 158.2591758599028948 i
 0.10410257317651053723 + 158.2591758599028948 i
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  • $\begingroup$ @Thanks Joro. The master of root finding at work! Think these are indeed exceptions for the first part of the conjecture, but what about the second part? Could there be a zero on $\Re(s)=\frac12$ for $|\zeta(s) + \dfrac{1}{\zeta(1-s)}|-2$ ? $\endgroup$
    – Agno
    Commented Nov 28, 2012 at 9:06

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