Is Riemann Zeta Function symmetrical about the real axis?

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? –  Rahul Dec 26 '10 at 14:33
Well, $k^{-s}$ has mirror symmetry about the real axis... –  Ｊ. Ｍ. Dec 26 '10 at 14:34
rpg: You live, you learn... ;) –  Ｊ. Ｍ. Dec 26 '10 at 14:49
$k^{-s}$ has mirror symmetry about the real axis? Not where I'm from, it doesn't. –  TonyK Dec 26 '10 at 20:34
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? –  TonyK Dec 27 '10 at 7:30
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.