question on Riemann $\zeta(s)$

Question

I have a question that is troubling me.

From the functional equation of $\zeta(s)$, can we not conclude that both $\zeta(s)$ and $\zeta(1-s)$ have the same non-trivial zeros (differing at most in the sense that roots of one is the complex conjugate of the other and vice versa) and if this were true doesn't this imply RH?

Thanks

Answer 1

Consider $$f(z) = z(1-z) - i$$

This satisfies $$f(z) = f(1-z)$$

but none of the zeroes have $\mathrm{Re}(z) = \dfrac{1}{2}$