is it true that $\zeta(\frac{1}{2} + bi) = 0 \implies \zeta(a + bi) \neq 0$ for $0 < a < 1$, $a\neq \frac{1}{2}$?

Question

I was wondering if a zero on the critical line implies no zero for the zeta function anywhere else in the critical strip for the same ordinate and vice-versa? I don't know if there is a proof for this.

That is does,

$\zeta(\frac{1}{2} + bi) = 0 \implies \zeta(a + bi) \neq 0$ for $0 < a <\frac{1}{2}$ and $\frac{1}{2} < a < 1$

Thank you

Answer 1

No. In fact for any negative even integer $$ \zeta(-2n) = 0\ \forall\ n \in \mathbb{N} $$

Answer 2

No, it is not. I'll try with contraposition. Your statement is if $\zeta(\frac12+bi)=0$, then for all $\frac12\ne a\in \Bbb R, \zeta(a+bi)\ne0$. So, its contraposition is if there exists $\frac12\ne a\in \Bbb R$ such that $\zeta(a+bi)=0$, then $\zeta(\frac12+bi)\ne0$.

The Riemann zeta holds this functional equation: $$ \zeta(x)=2^x\pi^{x-1}\sin\frac{\pi x}2\Gamma(1-x)\zeta(1-x) $$ Substitute $x$ as $-2n$ $(0<n\in\Bbb Z)$, then the term $\sin\frac{\pi x}2$ goes to $0$, so $\zeta(-2n)=\zeta(-2n+0i)=0$.

But $\zeta(\frac12+0i)=-1.46035\cdots\ne0$, so the contraposition is false.

So your statement goes to false.