# Zeros of Riemann zeta function, functional equation and Euler product

1. Jul 6, 2017

### binbagsss

1. The problem statement, all variables and given/known data

Question

Use the functional equation to show that for :

a) $k \in Z^+$ that $\zeta (-2k)=0$
b) Use the functional equation and the euler product to show that these are the only zeros of $\zeta(s)$ for $Re(s)<0$ . And conclude that the other zeros are all located in the critical strip: $0\leq Re(s) \leq 1$ . Show that these are symmetric about $s=1/2$

2. Relevant equations

Euler product: $\zeta(s)=\Pi^{p}\frac{1}{1-p^{-s}}$ defined for $Re(s)>1$

Functional equation:

$\zeta(s)=\chi(s)\zeta(1-s)$
where $\chi(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2} \Gamma(1-s)$

Also have $Z(s)=\pi^{\frac{-s}{2}} \Gamma(\frac{s}{2}) \zeta(s)$ which we know has simple poles at $s=0, 1$

So from this we can see that the $\Gamma (s)$ that gave poles for $Z(s)$ gives arise to the zeros of $\zeta(s)$ at $s=-2k$ so that's the trivial zeros done.

3. The attempt at a solution

From the Euler product define for $Re(s) > 1$ we can see that $\zeta (s)$ does not vanish for $Re(s) >1$.
I think to make the rest of the conclusions about the critical strip and being symmetrically distributed about $Re(s)=1/2$ I need to use the functional equation. But I'm not sure what to do...

I want to look where it is positive and negative I guess. But with $sin$ and $\Gamma$ which are positive and negative at different ranges of $s$ I'm not really sure what to do.. any hint greatly appreciated.

2. Jul 11, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.