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From wikipedia, http://en.wikipedia.org/wiki/Riemann_zeta_function

"Furthermore, the fact that $\zeta(s) = \zeta(s^*)^*$ for all complex s ≠ 1 ($s^*$ indicating complex conjugation) implies that the zeros of the Riemann zeta function are symmetric about the real axis."

I know the zeros are symmetrical. But what about the other values of $\zeta(s)$? My main aim is to find out:

Is $\zeta(s)$ symmetrical about the real axis for all $\Re(s) > 1$ ?

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    Isn't the answer to your question contained in the very sentence you quoted?2010-12-26
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    Well, $k^{-s}$ has mirror symmetry about the real axis...2010-12-26
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    ...and to add to Rahul's comment, that's why $\overline{f(\bar{s})}=f(s)$ is called a mirror symmetry identity.2010-12-26
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    @Rahul I wanted to find a proof of that fact. I should have said that in my question.2010-12-26
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    Ah. In that case, J.M.'s first comment should do, yes?2010-12-26
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    Yup... :) Thanks J.M. (I am so stupid I miss these simple steps...)2010-12-26
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    rpg: You live, you learn... ;)2010-12-26
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    $k^{-s}$ has mirror symmetry about the real axis? Not where I'm from, it doesn't.2010-12-26
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    @Tony, well, $\exp(\bar{s}\ln(k))=\overline{\exp(s\ln(k))}$ for $k > 0$ ... unless you intended to say something else.2010-12-27
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    To me, if $f$ has mirror symmetry about the real axis, then $f(\bar s) = f(s)$, not $f(\bar s) = \overline{f(s)}$. Am I wrong?2010-12-27
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    @Tony: I'm using [this way](http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/04/02/01/), though if you have convincing reasons to not call it a "mirror symmetry" relation, then sure, I'll stop using the term. :)2010-12-27
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    Now you've gone from 'has mirror symmetry' to 'has a mirror symmetry relation'. But it seems to me that the OP's Question is unambiguous: 'Is $\zeta(s)$ symmetrical about the real axis?' No it's not.2010-12-27

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Let $f$ be a holomorphic function with $f(\overline{z}) = f(z)$, then $f$ is necessarily a real constant function, so the most you can ask for is $\zeta( \overline{z}) = \overline{\zeta(z)}$, since the Riemann zeta function is not a constant function.