# {R. Zeta fn. zeros} = {Dir. eta fn. zeros} for 0<Re(s)<1?

1. Dec 23, 2013

if Î·(s) is the Dirichlet eta function and Z(s) is the Riemann zeta function, is it true that for s such that 0 < Re(s) < 1, Î·(s) =0 iff Z(s) = 0?

Ah, sorry for the Z for zeta, but for some reason the sigma symbol is not appearing in my toolbar.

2. Dec 24, 2013

### Newu

I think it is correct.Because we have
$$\eta (s) \equiv \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^s} = (1-2^{1-s})\zeta (s).$$
And that "The Riemann hypothesis is equivalent to the statement that all the zeros of the Dirichlet eta function $\eta(s)$ (a.k.a. the alternating zeta function) falling in the critical strip $0<\mathbf{R}<1$ lie on the critical line $\mathbf{R}=\frac{1}{2}$."http://mathworld.wolfram.com/RiemannHypothesis.html

3. Dec 24, 2013

Thanks, Newu. Since I believe I read that all the non-trivial zeros of the zeta function have to lie in the interval (0,1) anyway, then it would seem to be easier to work with the Dirichlet eta function than with the zeta function in order to decide the Riemann hypothesis, no?

4. Dec 24, 2013

### Newu

Maybe,they are equally hard,otherwise,just as you say,mathematicians would work with the Dirichlet function.
Merry Christmas,Merry Christmas Eve :-）

5. Dec 25, 2013

Merry Christmas also to you, Newu.
Just to check: all zeros of the factor (1-21-s) have Re(s) =1?

Last edited: Dec 25, 2013