Zeros of Riemann zeta function, functional equation and Euler product

[binbagsss]

Homework Statement

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 $s=1/2$

Homework 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.

The Attempt at a Solution

[/B]
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.

 

[mighty2000]

Thank you for your question. I am a scientist and I would be happy to help you with your problem. Let's start by looking at part a) of the question.

a) Using the functional equation, we can rewrite $\zeta(-2k)$ as $\chi(-2k)\zeta(1-(-2k))$. Since $k \in Z^+$, we know that $-2k$ is an even negative integer. Therefore, $\chi(-2k)=0$, as it is defined as $2^s\pi^{s-1}\sin(\frac{\pi s}{2} \Gamma(1-s))$. This means that $\zeta(-2k)=0$, as required.

Now, let's move on to part b) of the question. We can use the functional equation and the Euler product to show that the only zeros of $\zeta(s)$ for $Re(s)<0$ are located at $s=-2k$, where $k \in Z^+$.

First, let's look at the Euler product. We know that it is defined for $Re(s)>1$, but we can extend it to the region $0<Re(s)<1$ using analytic continuation. This means that the product is still valid for all values of $s$ in the critical strip $0\leq Re(s) \leq 1$.

Next, using the functional equation, we can rewrite $\zeta(s)$ as $\chi(s)\zeta(1-s)$. Since we are interested in the zeros of $\zeta(s)$ for $Re(s)<0$, we can write $\chi(s)\zeta(1-s)=\chi(s)\zeta(-s)$. From part a), we know that $\zeta(-s)=0$ for $Re(s)>0$, and therefore $\chi(s)\zeta(-s)=0$ for $Re(s)>0$.

This means that the only possible zeros of $\zeta(s)$ in the critical strip are located at $s=-2k$, where $k \in Z^+$. These are the trivial zeros that you have already mentioned.

Now, for the conclusion that the other zeros are all located in the critical strip, we