Proof of Symmetry of zeros in critiical strip

Question

How can we prove that zeros in critical strip are symmetric respect line $\sigma=\frac{1}{2}$ and respect real axis?

Answer 1

The Riemann zeta function satisfies two symmetry relations.

First, as an analytic function which is real-valued on the real line, $\zeta(\overline{s})=\overline{\zeta(s)}$; conjugating the argument conjugates the function. If $\zeta(s)=0$, $\zeta(\overline{s})=0$ and vice versa. That's the symmetry with respect to the real axis we wanted.

Second, we have Riemann's functional equation $$\zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$$ The exponentials in there are never zero or infinite. The $\sin$ is zero only at even integers, and never infinite. The $\Gamma$ function is never zero, and infinite only at $0$ and negative integers. So then, in the critical strip with real parts in $(0,1)$, $\zeta(s)=0$ if $\zeta(1-s)=0$ and vice versa. To get from $x+iy$ to $(1-x)+iy$, we combine this with the symmetry across the line; a zero at $x+iy$ implies one at $(1-x)-iy$, which we then reflect across the axis to get one at $(1-x)+iy$.