Building on this question: Zeros of $\zeta(s) \pm \zeta(1-s)$, I experimented further with:

$$\zeta(s) \pm \zeta(\overline s)$$

Assuming $s=\sigma + ti$, I observed that this function also has many "semi-trivial" as well as "non-trivial" zeros for each $\sigma$. Furthermore these "non-trivial" zeros all seem to reside very close to the Riemann non trivial zeros at $\sigma=\frac12$.

However, what I found curious is that only when $\frac12 < \sigma < 1$ the function:

$$\zeta(s) + \zeta(\overline s)$$

suddenly loses all of its "non-trivial" zeros (i.e. the ones near the Riemann zeros), whilst still retaining all of its "semi-trivial" zeros (they disappear when $\sigma >1$). Is there a logical explanation or even proof for this?

P.S.:

In an attempt to find out more, I used the alternating zeta-function $\eta(s)$ and rewrote it as:

$$\eta(s) - \eta(\overline s) =\displaystyle 2i \sum _{n=1}^{\infty } \frac{e^{\pi i n} \sin(t \ln(n))}{n^\sigma}$$

and

$$\eta(s) + \eta(\overline s) =\displaystyle 2i \sum _{n=1}^{\infty } \frac{e^{\pi i n} \cos(t \ln(n))}{n^\sigma}$$

These functions look very symmetrical, but it seems that the denominator $n^\sigma$ drives the infinite alternating sum of the cosines to always positive near the non-trivial zeros whilst keeping the semi-trivial ones intact, when $\sigma > \frac12$. Could these functions be rewritten as an integral?