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!