What is the proof for any non trivial zero? [closed]

Question

There are many known nontrivial zeros of the Riemann Zeta function, but I have never seen proof that any of them actually resolve to zero. The trivial zeros make sense because there is a more complicated equation which represents the zeta function and in that equation all negative even integers resolve to 0.

Basically, is there proof that 14... times i plus 1/2 actually resolves to 0 in the zeta function? How do you know? What is the proof?

Is there proof that the non trivial zeros are actually zeros? If so, where is the proof? What is the evidence? Can anybody give one example???

Answer 1

For any function $f$ holomorphic on some domain $D$, the integral $$ \frac{1}{2\pi i} \int_{\partial D} \frac{df}{f} $$ is a nonnegative integer, and counts the number of zeros of $f$ in $D$. So if you can compute this integral to good enough precision, you know the exact number of zeroes in $D$. By taking $D$ a small enough disc around a suspected zero, you have a proof that there is actually a zero there.

Moreover, if $z$ is a zero of $\zeta$ such that $0<\Re(z)<1$, then $1-z$, $\overline{z}$, $1-\overline{z}$, are also zeroes in the same band (because of symmetries satisfied by the Zeta function). If there were a zero $z$ with $0<\Re(z)<1$, $\Im(z)=a$ and $\Re(z) \ne 1/2$, then $1-\overline{z}$ would be another zero satisfying the same conditions. For every small enough rectangle $R_\epsilon$ with vertices $a \pm i\epsilon$ and $1+a \pm i\epsilon$, $$ \frac{1}{2\pi i} \int_{\partial R_\epsilon} \frac{df}{f} $$ would be equal to $2$, and not to $1$.