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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

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  • $\begingroup$ By $\zeta(a,b),$ you mean $\zeta(a+bi)$? $\endgroup$ Jan 13, 2022 at 23:53
  • $\begingroup$ Yes. Sorry about that. $\endgroup$
    – sku
    Jan 13, 2022 at 23:58
  • $\begingroup$ it is true. I think I have the proof on my notebook, give me a sec and I will look for it brb $\endgroup$ Jan 14, 2022 at 0:30
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    $\begingroup$ I would be surprised if it is proven. @hellofriends $\endgroup$
    – reuns
    Jan 14, 2022 at 4:45

2 Answers 2

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

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    $\begingroup$ i am interested in $a$ inside the critical strip. $\endgroup$
    – sku
    Jan 14, 2022 at 0:23
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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.

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  • $\begingroup$ i am interested in $a$ inside the critical strip. $\endgroup$
    – sku
    Jan 14, 2022 at 0:23
  • $\begingroup$ Then you'ld rather edit the title from $a\ne\frac12$ to $0<a<1, a\ne\frac12$. I'll try to answer about critical strip... $\endgroup$
    – MH.Lee
    Jan 14, 2022 at 0:25

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