1
$\begingroup$

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 $1/n^s$?

$\endgroup$
  • $\begingroup$ 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
2
+100
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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