Numerical evidence suggests that all complex zeros (real ones exist as well) of:

$$\frac{\zeta'}{\zeta}(s) \pm \frac{\zeta'}{\zeta}(1-s)$$

reside on the critical line with $\Re(s)=\frac12$.

I made some progress by taking:

(1) $\zeta(s):=\chi(1-s)\,\zeta(1-s), \, \, \chi(s)= \Gamma \left( s \right) \cos \left(\frac12\pi s \right) 2 \left( 2\pi \right) ^{-s}$

(2) $\zeta(s):= -\dfrac{\zeta'(1-s)+\chi(s)\,\zeta'(s)}{\chi'(s)}$ (derived from *Apostol's paper found here*).

and then the formulae can be rewritten into:

\begin{align*} \frac{\zeta'}{\zeta}(s) + \frac{\zeta'}{\zeta}(1-s) &= -\dfrac{\chi'}{\chi}(s) \\ \\ \frac{\zeta'}{\zeta}(s) - \frac{\zeta'}{\zeta}(1-s) &= -\dfrac{\chi'}{\chi}(s) \cdot \dfrac{\chi(s)\,\zeta'(s)-\zeta'(1-s)}{\chi(s)\,\zeta'(s)+\zeta'(1-s)} \\ \end{align*}

The (+) version only has a single pair of zeros at $\frac12 \pm 6.2898359888369027796...$ and shows a monotonically increasing absolute value from that point onwards (note that the expected poles at the non-trivial zeros $\rho$ are all annihilated by $\zeta'(1-s)+\chi(s)\,\zeta'(s)$) that also induces the $\rho$s).

The (-) version does have a pole at each $\rho$, however these appear to be always separated by a single new zero that apparently always resides on the critical line. Could the latter be proved?

Thanks.

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