Some of my computations [here](https://www.wolframalpha.com/input/?i=zeta(s)(zeta(1-s)),+s%3D0.5%2B500i) 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 $1/2$.



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> > **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 ?
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> **Note:** $ \zeta(s)$ is the Riemann zeta function with $s$ is a complex variable .
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Is the imaginary part of $\displaystyle\ \zeta(s)\zeta(1-s)=0$ for $\operatorname{Re}(s)=\frac{1}{2}$
$\overline{\zeta(s)\zeta(1-s)} = \zeta(\overline{s})\zeta(\overline{1-s}) = \zeta(1-s)\zeta(s)$


($1-s=\overline{s}$ 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.


