Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\Re(s) > 1$ ?

share|improve this question
6  
Isn't the answer to your question contained in the very sentence you quoted? –  Rahul Dec 26 '10 at 14:33
4  
Well, $k^{-s}$ has mirror symmetry about the real axis... –  J. M. Dec 26 '10 at 14:34
1  
rpg: You live, you learn... ;) –  J. M. Dec 26 '10 at 14:49
1  
$k^{-s}$ has mirror symmetry about the real axis? Not where I'm from, it doesn't. –  TonyK Dec 26 '10 at 20:34
1  
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
show 7 more comments

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.