-2

I can prove that there are no zeros of $\zeta$ in the region Re$(z)>1$. This follows from

$$\zeta(z)\prod_{p} (1-p^{-z})=1$$

I assume you all know the proof, it is pretty easy so I won't type it. My question is this: How do we know that there are no NON-TRIVIAL zeros in the left complex half-plane where Re$(z)<0$? If there is a theorem, the name of the theorem will be very helpful. I've seen that this follows from the Theorem of Hadamard and de la Vallee-Poussin who showed that there are no zeros on the line Re$(z)=1$, but I am having trouble seeing how it follows about Re$(z)<0$. Thanks!

hodop smith
  • 192

2 Answers2

5

As mentioned, there are trivial zeros at $\zeta (-2n)$ for all positive integers $n$. But besides these trivial zeros, we know that there are no other zeros to the left of the critical strip by the functional equation $\pi ^{\frac{s}{2}}\Gamma (\frac{s}{2})\zeta (s)=\pi ^{-\frac{1-s}{2}}\Gamma (\frac{1-s}{2})\zeta (1-s)$, since we already know there are no zeros to the right of the critical strip. If there were to be a nontrivial zero in the half plane $\Re (s)<0$, then there would be a corresponding zero in the half plane $\Re (s)>1$.

3

We use the functional equation which has the form $$\zeta(1-s)=F(s)\zeta(s)$$ where $F(s)$ is an explicit "fudge factor". Then for $\text{Re}\,s>1$, $\zeta(1-s)\ne0$ except if $F$ has a zero at $s$. So $\zeta$ is nonzero on the half-plane $\text{Re}\,s<0$ except if $F(1-s)=0$, which does occur for the "trivial" zeros, $s=-2n$, for $n\in\Bbb N$.

Angina Seng
  • 158,341