Why is $\zeta(s)\neq0$ for $\operatorname{Re}(s)=0$?

Question

I have a question concerning the Riemann zeta function for a project I've been working on. Why is it that $\zeta(s)\neq0$ for $\operatorname{Re}(s)=0$ (that is, there are no non-trivial zeroes of the zeta function lying on this line)? My guess would be that this follows from the zeta functional equation: $$ \zeta(s)=2(2\pi)^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s) $$ and the symmetry of zeroes about the line $\operatorname{Re}(s)=1/2$. More precisely, I was aware that for $0<\operatorname{Re}(s)<1$, the functional equation shows us that if $s$ is a zero then $1-s$ is also a zero. I wondered then if we can further say that this holds for $0\leq\operatorname{Re}(s)\leq1$, where $s\neq1$? This would mean that, by reflection, as $\zeta(0)\neq0$, we can deduce that the line $\operatorname{Re}(s)=0$ contains no zeroes. A confirmation or explanation would be much appreciated.

Answer 1

Using the functional equation shows that if if Re(s) = 0 for some Im(s) = t is a zero, then Re(s) = 1 is a zero for the same t. This is not possible since there are no zeros on the line Re(s) = 1 which is a pole for Zeta. There is also a zero free zone asymptotic to Re(s) = 1 in the critical strip.

Answer 2

It is correct to say that a zero at $s$ yields a zero at $1-s$. More specifically, if we call $$\xi(s) = \pi^{\frac{s}{2}} \Gamma(\tfrac{s}{2})\zeta(s),$$ then $\xi(s)$ has understandable meromorphic properties and the functional equation $s \mapsto 1-s$ is true for all $s$, and one can infer information about the zeroes of the zeta function from the poles of the gamma function. Since the gamma function is never zero, at any point where $\xi(s) = 0$, we must have $\zeta(s) = 0$. Further, since $\Gamma(s)$ has simple poles at $s = -n$ for $n$ a positive integer, then there must be zeroes of the zeta function at $-2n$.

As there are no zeroes of the zeta function with $\text{Re} (s) = 1$, there are no zeroes with $\text{Re} (s) = 0$. The former is a conclusion often proved in demonstrating the prime number theorem (and is in fact one of the two main hurdles in the proof of the prime number theorem; the other is how to translate from this fact to a sharp asymptotic, which requires either a zero-free region or some delicate Laplace transform arguments).