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For $s=\sigma+it$, the Riemann Zeta series

$$\zeta(s)=\sum \frac{1}{n^s}$$

valid for $\sigma>1$, and the extended version

$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum\frac{(-1)^{n+1}}{n^{s}}$$

valid for $\sigma>0$, and both symmetric about the real axis.

That is, $\zeta(\sigma+it)=\zeta(\sigma-it)$.

Question: How can I prove the functions are symmetric about the real-axis, that is, $\zeta(\sigma+it)=\zeta(\sigma-it)$

Note - the suggested question doesn't answer this question sufficiently clearly: Is Riemann Zeta Function symmetrical about the real axis?

Penelope
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    This is not true; rz is conjugate invariant so $\overline \zeta(s)=\zeta(\bar s)$ – Conrad Aug 29 '21 at 00:44
  • @Conrad thanks, I have been looking a visualisations of the magnitude $|\zeta(s)|$ which hides this. Can you recommend a proof/argument that shows $\overline \zeta(s)=\zeta(\bar s)$? – Penelope Aug 29 '21 at 01:06
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    Just use the definition for $\Re s > 1$ and analytic continuation for the rest (using that $\bar \zeta (\bar s)=\zeta(s)$ on $\Re s>1$ and both sides are analytic, hence equality holds everywhere) – Conrad Aug 29 '21 at 01:21
  • Thanks @Conrad - it took me some time with a pen and paper to confirm that each term of the series $1/n^\overline{s}=\overline{1/n^s}$. This wasn't obvious to me as in general $f(\overline{s}) \neq \overline{f(s)}$. – Penelope Aug 30 '21 at 22:00
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    Dirichlet and Taylor series (centered at a real) with real coefficients are clearly conjugate invariant where they converge absolutely and then any analytic continuation of the functions defined by them is too; conversely it is not hard to prove that a conjugate invariant Dirichlet or Taylor series (centered at zero or another real of course) that converges on an open non empty set must have real coefficients as in either case identically zero implies zero coefficients and $\bar f(\bar s)-f(s)=0$ when $f$ is conjugate invariant, while in both cases the coefficients of that are $\bar a_n-a_n$ – Conrad Aug 30 '21 at 22:17
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    Actually in the comment above one needs the domain of the analytic continuation to be conjugate invariant too as there are easy example of analytic functions that have real coefficients Taylor series centered at some reals but not at others (eg $\sqrt z$ defined on a connected tubular neighborhood of $(1,2) \cup (-2,-1)$ which of course cannot be extended analytically to any conjugate invariant domain – Conrad Aug 30 '21 at 22:27
  • hi @Conrad - apologies for being slow on this (self taught), is there anywhere I can read about the principle which allows me to say that the conjugate invariant applies over all $s$ except at singularities? I have read in several places that if it applies in some domain, then it must apply everywhere but I don't know the real reason why. – Penelope Sep 11 '21 at 13:53
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    read about analytic continuation (if $f,g$ analytic on some domain and $f=g$ on a set with accumulation points inside the domain - eg an interval included in the domain - then $f=g$ on the full domain; as long as $z,\bar z$ are both in the domain and $f$ is real on a real interval, one can apply this to $f, g=\bar f(\bar z)$) – Conrad Sep 11 '21 at 13:58

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