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 $1/n^s$?

  • $\begingroup$ 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 $\endgroup$ – DonAntonio Dec 14 '13 at 19:10
  • $\begingroup$ @DonAntonio The equation is not wrong, it is just usually written $$\zeta(s)=\prod_p\left(1-\frac{1}{p^s}\right)^{-1}.$$ $\endgroup$ – anon Dec 14 '13 at 19:12
  • $\begingroup$ Oh, I missed there is no $\;p^{-s}\;$ but $\;p^s\;$ ...ok, thanks. $\endgroup$ – DonAntonio Dec 14 '13 at 19:14

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.

  • $\begingroup$ 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. $\endgroup$ – PJ Miller Dec 14 '13 at 19:14
  • $\begingroup$ @PJMiller locally, a meromorphic function is given by a Laurent series; what happens when you conjugate the argument? $\endgroup$ – anon Dec 14 '13 at 19:18
  • $\begingroup$ But the Laurent series only holds in the neighborhood of a certain point $s$, so it doesn't hold near $\bar{s}$, does it? $\endgroup$ – PJ Miller Dec 14 '13 at 19:20
  • $\begingroup$ @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 $\zeta$. $\endgroup$ – anon Dec 14 '13 at 20:07
  • $\begingroup$ Could you please give more detail on that? It's really not obvious why it should be. $\endgroup$ – PJ Miller Dec 14 '13 at 20:09

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