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Some of my computations here showed to me that the imaginary part of $\displaystyle\ \zeta(s)\zeta(1-s)=0$ for $\operatorname{Re}(s)=\frac{1}{2}$, really i w'd like to know if there is any paper showed this result that is true for all s with a real part equal's $1/2$.

My question here is: Is imaginary part of ($\displaystyle\ \zeta(s)\zeta(1-s))=0$ for $\operatorname{Re}(s) =\frac{1}{2}$ and is there any paper discussed this ?

Note: $ \zeta(s)$ is the Riemann zeta function with $s$ is a complex variable .

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    $\begingroup$ Although the expert answer to the question is "obviously X is true for long-known reasons", the not-quite-experts may be blind-sided by such questions, depending how they approach them... e.g., numerically. Given the small space that it takes to archive such question-and-immediate-answer, it might be worthwhile to keep such things around. As in the fallacy of "it's so clear to experts that they never explain it". $\endgroup$
    – paul garrett
    Commented Jan 25, 2017 at 1:03
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    $\begingroup$ As a rule, I don't support postings with "i w'd" on this site. $\endgroup$
    – Franz Lemmermeyer
    Commented Jan 25, 2017 at 13:25

1 Answer 1

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$\overline{\zeta(s)\zeta(1-s)} = \zeta(\overline{s})\zeta(\overline{1-s}) = \zeta(1-s)\zeta(s)$

($1-s=\overline{s}$ for $Re(s)=1/2$)

So this is indeed a purely real number. $\zeta(\overline{s}) = \overline{\zeta(s)}$ holds because $\overline{\zeta(\overline{s})}-\zeta(s)$ is analytic and zero for real $s>1$, hence zero everywhere.

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