Is there a relatively simple proof for the following related identities (shown true in Mathematica only) involving the the analytically continued Riemann Zeta Function, $\zeta(s)$?
$$\zeta (s) \zeta (1-s)=\left(\zeta \left(s^*\right) \zeta \left(1-s^*\right)\right)^*\tag{1}$$
and
$$\zeta (s) \zeta \left(1-s^*\right)=\left(\zeta \left(s^*\right) \zeta (1-s)\right)^*\tag{2}$$
$s$ is complex and $s^*$ is the complex conjugate of $s$.
I presume any proof will involve some variation of Riemann's Functional Equation,
$$\zeta(s)=2^s \pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s)$$
but this does not seem straight forward and I can't see how.
If we now assume $s=\frac{1}{2}+i z$ and $s^*=\frac{1}{2}-i z$ then $\zeta(s)\zeta(1-s)=(\zeta(s^*)\zeta(1-s^*))^*$ seems to always be a real product whatever the value of $z$, but $\zeta(s)\zeta(1-s^*)=(\zeta(s^*)\zeta(1-s))^*$ seems to only have a real product when both the Zeta Functions involved are zero or $z=0$. Assuming this is correct, does this difference follow from the proof above?
Best Answer
This is actually much simpler than it appears: by the Schwarz reflection principle, any analytic function $f(z)$ that is real on the real axis satisfies $f(\overline z) = \overline{f(z)}$ and thus $f(z) = \overline{f(\overline z)}$.