This property is rather elementary, and not at all specific to $\zeta$, so I am wondering if it has any value in studying the zeros of the Riemann zeta function in the critical strip. It is a well known result? I can provide a proof sketch if you are interested, and it has been checked numerically.


If $\zeta(s)=0$, with $s=\sigma +it$ and $0<\sigma<1$ then for all real $\theta$, we have


$$\sum\_{n=1}^{\infty}(-1)^{n+1}\frac{\cos(\theta+t\log n)}{n^\sigma}=0.$$


This would be true even if by chance, one of the zeroes is outside the critical line $\sigma=\frac{1}{2}$. In particular let $t\_0$ be the imaginary part of a zero of $\zeta(s)$. Then in order to find all $\sigma$'s such that $\zeta(\sigma +it\_0)=0$, we only need to focus on the $\sigma$'s that satisfy the following equation for every $\theta$:


$$\sum\_{n=1}^{\infty}(-1)^{n+1}\frac{\cos(\theta+t\_0\log n)}{n^\sigma}=0.$$


Of course the Riemann Hypothesis (RH) implies that $\sigma$ must be equal to $\frac{1}{2}$, but my assertion is true even if RH is not true. So if you prove that my equality can only happen if $\sigma=\frac{1}{2}$, then you would have proved RH. I have several strong reasons to believe that my equality does not lead to a proof of RH, yet I am wondering if it has any value.



On some property of the zeros of $\zeta(s)$ in the complex plane
That looks like just one of the two components of a "rotated" Dirichlet Eta function (sometimes called Alternate Zeta function):
$$e^{i\theta} \; \eta (s)= e^{i\theta} \; \sum \_{n=1}^{\infty}{\frac {(-1)^{n+1}}{n^{s}}}$$
It cannot hence help, as it is always possible to find a rotation angle $\theta$ that will bring to zero one of the two components. True that the non trivial zeros of $\eta$ coincide with those of the Riemann Zeta function $\zeta$. But the difficulty is of course that zeros of both said functions require by definition that both components be zero.


Sure enough, if $\eta(s) = 0$ then it would be so also under any rotation $\theta$.


Conversely, if any one of $\eta(s)$ two components remains $0$ under any rotation $\theta$, then $\eta(s)=0$.


