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If $\zeta(x)=a+ib$ and $\zeta(y)=a-ib$, is there a single equation that relates $\zeta(x)$ to $\zeta(y)$?

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Indeed, it is true that any meromorphic function $f(z)$ that is real on the real axis satisfies $f(\bar z) = \overline{f(z)}$. This fact is a combination of the Schwarz reflection principle and the uniqueness of analytic continuations.

(Even more pedantically: all that is required is that $f(z)$ is real on a subset of the real axis with a limit point inside the domain of analyticity. I mention this because the simplest way to see that $\zeta(s)$ is real on the real axis is via the Dirichlet series $\zeta(s) = \sum_{n\in\Bbb N} n^{-s}$, which is valid on the interval $(1,\infty)$.)

Greg Martin
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  • Hello, why all that is required is that $f(z)$ is real on a subset of the real axis with a limit point inside the domain of analyticity? – one potato two potato Nov 26 '22 at 13:10
  • I don't know if there's any good answer to that "why" question other than to read the proof of the implication and see that the assumption suffices. – Greg Martin Nov 26 '22 at 18:20
  • I know the reflection principle. By the series representation, $\zeta$ maps $(1,\infty)$ to $\Bbb R$. But this does not imply $\zeta$ is real on the real axis. My understanding of the statement you wrote is that 'If $A\subset\Bbb R$ is a nonempty open subset and a holomorphic $f$ maps $A$ to $\Bbb R$ then $f$ maps the whole $\Bbb R$ to $\Bbb R$ (Here, we assume $f$ is holomorphic on $\Bbb R$ except poles). I was wondering why this is true. – one potato two potato Nov 27 '22 at 04:46