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

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

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

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

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

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 $$-2n$$ $$(0, then the term $$\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.

• i am interested in $a$ inside the critical strip.
– sku
Jan 14, 2022 at 0:23
• 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... Jan 14, 2022 at 0:25