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I'm studying the Riemann $\zeta$-function, and I'm now finding out that, apparently, whenever $\rho$ is a zero of $\zeta$, $\overline{\rho}$ is also one. This follows easily from the identity $\zeta(\overline{s}) = \overline{\zeta(s)}$, which I saw on wikipedia and here: Is Riemann Zeta Function symmetrical about the real axis?.

However, how does one prove this identity?

Matija Sreckovic
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    Can you first prove it for $\mathrm{Re}; s > 1$ ? Then you will need to understand the analytic continuation involved for $\zeta(s)$. – GEdgar Feb 21 '18 at 02:04
  • Does $\overline{n^{s}} = n^{\overline{s}}$ hold? I'm not sure about that step. I'm trying to prove it for the sum $\sum_{n=1}^{\infty} \frac{1}{n^{s}}$, should I be trying to do it for the Euler product instead? – Matija Sreckovic Feb 21 '18 at 11:42
  • Yes, $\overline{n^{s}} = n^{\overline{s}}$ is a good first step. What is the definition of $n^s$ for positive integer $n$ and complex $s$ ?? – GEdgar Feb 21 '18 at 11:43
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    I was able to work it out for $\Re s > 1$, as $\overline{n^{s}} = e^{\sigma \ln{n}}(\cos{t \ln{n}} - i \sin{t \ln{n}}) = n^{\sigma}e^{-it\ln{n}} = n^{\sigma -it} = n^{\overline{s}}$. – Matija Sreckovic Feb 21 '18 at 11:47
  • In class, we learned about the analytic continuation through the Euler-Maclaurin summation formula. Is that approach applicable to proving the identity? – Matija Sreckovic Feb 21 '18 at 11:48
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    No, I would say what you need to use is that the analytic continuation exists. The two functions $$\zeta(s)\text{ and }\overline{\zeta(\overline{s})}$$ are both analytic on the connected set $\mathbb C \setminus {1}$, and they agree on a set with a limit point, therefore they agree everywhere. – GEdgar Feb 21 '18 at 11:51
  • Ah, I see, because of the identity theorem? – Matija Sreckovic Feb 21 '18 at 11:56

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