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Show that if $a$ is a zero of the zeta function in the critical strip, then so are $\bar{a}$, $1-a$, and $1-\bar{a}$.

The definition of $\zeta$ is $$\dfrac{1}{\zeta(s)}=\prod_p\left(1-\frac{1}{p^s}\right)$$

I don't see how to get the desired fact from this. Or perhaps we should use the definition as sum of $1/n^s$?

PJ Miller
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  • That definition of zeta is wrong: either you meant $;\zeta(s);$ or else a minus one exponent is missing in the product...and besides this it works for $;\text{Re}(s)>1;$ , so it can't be you're working with this for this problem – DonAntonio Dec 14 '13 at 19:10
  • @DonAntonio The equation is not wrong, it is just usually written $$\zeta(s)=\prod_p\left(1-\frac{1}{p^s}\right)^{-1}.$$ – anon Dec 14 '13 at 19:12
  • Oh, I missed there is no $;p^{-s};$ but $;p^s;$ ...ok, thanks. – DonAntonio Dec 14 '13 at 19:14

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That is only the definition of $\zeta(s)$ for ${\rm Re}(s)>1$. The Euler product does not converge if the real part of $s$ is $\le1$*. Rather, $\zeta(s)$ is defined by the analytic continuation of the $p$-series $\sum n^{-s}$ (or I guess equivalently the Euler product $\prod (1-p^{-s})^{-1}$ if you want) to the rest of the complex plane.

This analytic continuation is achieved explicitly via the functional equation. You can conclude that $s$ is a zero iff $\bar{s}$ is a zero since $\zeta(\bar{s})=\overline{\zeta(s)}$ (as I pointed out in the comments, this follows from the series definition $\sum n^{-s}$; how does conjugation affect each term of this?) and conclude that $s$ is a zero iff $1-s$ is a zero using the functional equation.

*Actually I think the Euler product might converge for some $s$ with real part $1$, IIRC.

anon
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  • Thanks, I see now. The only thing I don't know is your conjugate equation for meromorphic functions. Let me think about it a bit. If I don't get it, maybe I'll start another thread for that. – PJ Miller Dec 14 '13 at 19:14
  • @PJMiller locally, a meromorphic function is given by a Laurent series; what happens when you conjugate the argument? – anon Dec 14 '13 at 19:18
  • But the Laurent series only holds in the neighborhood of a certain point $s$, so it doesn't hold near $\bar{s}$, does it? – PJ Miller Dec 14 '13 at 19:20
  • @PJMiller You're right, not in general. A different tact, then. Show $\zeta(\bar{s})=\overline{\zeta(s)}$ where the series $\sum n^{-s}$ converges, so it must be true on the whole domain of $\zeta$. – anon Dec 14 '13 at 20:07
  • Could you please give more detail on that? It's really not obvious why it should be. – PJ Miller Dec 14 '13 at 20:09
  • @PJ same idea as with the Laurent series hint; look at how conjugation affects each term – anon Dec 14 '13 at 20:14
  • Based on OP's saying "the solution by anon is not correct, as pointed out in the comments," I don't think OP understands my previous comment about conjugation and $n^{-s}$. A hint: show $a^\bar{s}=\overline{a^s}$ for any appropriate $a$ by arguing it for $a=e$ in particular (use the power series definition, and the fact that $\bar{s}^n=\overline{s^n}$). Note though: the fact that Laurent series are not globally convergent is not a big obstacle to proving this for all meromorphic functions; one only needs it to converge around a single point on the real line (which is guaranteed). – anon Dec 17 '13 at 14:01
  • Although I do suppose we need the fact that the coefficients are all real numbers, which is nontrivial, so it's better to use this $n^{-s}$ approach. – anon Dec 17 '13 at 14:10