-1

Ror Riemann's zeta function if one proves a relation in the domain $\operatorname{Re}(s)>1$ will this be sufficient proof that it is satisfied in all the z domain

for example if one proves $\zeta(s)=\overline{\zeta\left(\overline s\right)}$ from the original function defined in $\operatorname{Re}(s)>1$ does it require another proof for $\operatorname{Re}(s)<1$ or can we say that it is satisfied for all z domain and why?

José Carlos Santos
  • 427,504
Mina Maged
  • 3
  • It's not too easy for me to guess what you really mean, but for example $;\zeta(s)\neq0;$ for $;s\in\Bbb C,,\text{Re},s>1;$ , yet this is false in the whole domain... – DonAntonio Jun 08 '18 at 21:50

1 Answers1

2

The general question is too broad, but it is easy to answer in the case of the specific example that you mentioned. Since $\zeta(s)=\overline{\zeta\left(\overline s\right)}$ is an equality between analytic functions, then, by the identity theorem and because the domain of the $\zeta$ function is connected, if this identity holds when $\operatorname{Re}s>1$, it holds everywhere.

José Carlos Santos
  • 427,504
  • does this mean that for example if we solve ζ(s)=ζ(s¯) using the form of zeta function defined for re(s)>1 we will have solved it in the entire z domain? if so how come this solution does not include the non trivial zeros ? – Mina Maged Jun 09 '18 at 18:45
  • @MinaMaged Actually, what I did proves that $\zeta(s)=0\iff\zeta\left(\overline s\right)=0$.Since the trivial zeros are real, this provides no information about them. But it it tells us that we only need to look for non-trivial zeros with non-negative imaginary part. – José Carlos Santos Jun 09 '18 at 18:50