# A question of roots of riemann function

1. Aug 9, 2005

### eljose

let be the quotient:

$$Lim_{x->c}\frac{\zeta(1-x)}{\zeta(x)}$$
where x=c is a root of riemann function... then my question is if that limit is equal to exp(ik) with k any real constant...thanks... the limit is wehn x tends to c bieng c a root of riemann constant

2. Aug 9, 2005

### shmoe

If c is a trivial root, the answer is clearly no.

If c is a non-trivial root, if you use the functional equation the answer should be clear.

3. Aug 9, 2005

### matt grime

no, that can't be true, k surely cannot be *any* real constant? it can at best be *some* real constant.

4. Aug 9, 2005

### eljose

let,s suppose c is a trivial root i mean 0<re(c)<1 i have checked with the functional equation the limit $$Limx->c\frac{\zeta(1-x)}{zeta(x)}$$ my question is if we can express this limit as exp(ik) with k some real constant...

i have checked the functional equation for zeta if we call $$\zeta(1-s)=F(s)\zeta(s)$$ the only thinkg i have proved is that F^*(a+ib)=F(a-ib) (with * complex conjugate) but i don,t know if it will be true that the coefficient is equal to exp(ik).

5. Aug 9, 2005

### shmoe

if 0<Re(c)<1 it is a NON-trivial root.

$$\frac{\zeta(1-s)}{\zeta(s)}=\chi(1-s)$$, and I'm sure you know what $$\chi$$ is despite your continued use of "F".

So it depends on $$\chi(1-c)$$. If c is on the critical line, this has modulus 1. If c is off the critical line, this may have modulus 1 as well, or it may not.