# Two incremental sets of four zeros for $\zeta(s) \pm \zeta(\overline{1-s})$

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

I like to conjecture that it only becomes zero for (obviously) all the non-trivial zeros $\rho_n$, however then also for two sets of four roots $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$ that are symmetric around $s=\frac12$:

$$\zeta(\mu) + \zeta(\overline{1-\mu}) = 0 \rightarrow \mu_{+} = 6.023362... +5.506529...i$$

$$\zeta(\mu) - \zeta(\overline{1-\mu}) = 0 \rightarrow \mu_{-} = 8.990946... +4.510581...i$$

1) Is this a known result?

2) Since $\zeta(s)$ with $\Im(s)>0$ cannot become zero outside the critical strip, I wondered whether there might be a way to calculate these incremental roots directly from the reflection formula e.g. by firstly transforming $\zeta(s) \rightarrow \zeta(\overline{s})$ (but how?) and then taking $\zeta(\overline{s}) =\chi(1-\overline{s}) \zeta(1-\overline{s})$ and equating $\chi(1-\overline{s})$ to $1$ or $-1$?

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