TITLE: Are all complex zeros of $\dfrac{\zeta'}{\zeta}(s) \pm \dfrac{\zeta'}{\zeta}(1-s)$ on the critical line $\Re(s)=\frac12$?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: See the following papers which treat this and similar questions.

MR1986257 (2004c:11152) Reviewed 
Saidak, Filip(3-CALG-MS); Zvengrowski, Peter(3-CALG-MS)
On the modulus of the Riemann zeta function in the critical strip. (English summary) 
Math. Slovaca 53 (2003), no. 2, 145–172. 
MR3277049 Reviewed 
Matiyasevich, Yu.(RS-AOS2); Saidak, F.(1-NCG); Zvengrowski, P.(3-CALG-MS)
Horizontal monotonicity of the modulus of the zeta function, L-functions, and related functions. 
Acta Arith. 166 (2014), no. 2, 189–200.