# Are all complex zeros of $Li_s(i)\, + \, Li_{1-\overline{s}}\,(-i)$ equal to the $\rho$'s?

Take the well known square relationship for polylogarithms:

$$Li_s(z)\, + \, Li_{s}(-z)=2^{1-s}Li_s(z^2)$$

Assume $z=i$:

$$Li_s(i)\, + \, Li_{s}(-i)=2^{1-s}Li_s(-1)=-2^{1-s}\,\eta(s)$$

with $\eta(s)$ being the Dirichlet Eta function.

Then make the following tweak:

$$Li_s(i)\, + \, Li_{1-\overline{s}}\,(-i)$$

that is obviously still equal to $-2^{1-s}\,\eta(s)$ when $\Re(s)=\frac12$ and therefore, assuming RH, induces the non-trivial zeros $\rho$.

Questions:

1) Is there a closed form for $Li_s(i)\, + \, Li_{1-\overline{s}}\,(-i)$?

Using $Li_s(\pm\,i)=-2^{-s} \,\eta(s) \pm i\,\beta(s)$, gives: $-2^{-s}\,\eta(s) -2^{\overline{s}-1}\, \eta(1-\overline{s})+i\,\left(\beta(s)-\beta(1-\overline{s})\right)$. Even though both the Dirichlet $\eta$ and $\beta$ functions have reflective formula relating $s$ to $1-s$, further simplification is blocked since there are no known relations to $1-\overline{s}$.

2) Does the RH imply that the only complex zeros of $Li_s(i)\, + \, Li_{1-\overline{s}}\,(-i)$ must be the $\rho$'s?