- #1

binbagsss

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## 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.