This question loosely builds on [this one][1], however is a bit simpler and I found the results to be more robust.

It seems that all zeros in the critical strip $0 \lt \Re(s) < 1$ of:

$$Li_s(z)\, \pm \, Li_{1-s}(z)$$

reside on the critical line $\Re(s)=\frac12$ for all $z \le -1$.

Below is a graph that shows where this function (using $\pm = -$) vanishes for $s=\frac12 \pm t\,i$ at different values of $z$. The lines marked in red all lead to a non-trivial zero of $\zeta(s)$, since $Li_s(-1)=\eta(s)$, with $\eta(s)$ being the [Dirichlet Eta-function.][2] 

I have extended the graph towards $z \rightarrow 0$ to also show how some lines continue in that domain, however there clearly are zeros off the critical line when $z > -1$ (note: the lines coming from the left don't have a 'hard stop' at $z=-1$ and actually continue a short bit further to the right).

![enter image description here][3]

For the root finding process I used the following expression of the PolyLog function that I found to be evaluating much faster for higher values than the standard PolyLog function (in Maple):

$$Li_s(z) = \frac{\Gamma(1-s)}{(2\,\pi)^{1-s}} \left(i^{1-s}\,\zeta_H\left(1-s,\frac12+\frac{\ln(-z)}{2 \,\pi \, i}\right)+i^{s-1}\,\zeta_H\left(1-s,\frac12-\frac{\ln(-z)}{2 \,\pi \, i}\right)\right)$$

where $\zeta_H(s,q)$ is the [Hurwitz zeta function.][4]


**Questions:**

1) Is there a counterexample with a zero lying off the critical line, but in the critical strip for $z \le -1$?

2) When $z \rightarrow 0$ the function $Li_s(z)\, - \, Li_{1-s}(z)$ vanishes at $s=\frac12 \pm \frac{k \, \pi \, i}{\ln(2)}$ with $k=0,1,2,3, \dots$, however I failed to derive this apparently trivial result from the known formulae. Could this be proven?

Thanks.

  [1]: http://mathoverflow.net/questions/177201/are-all-complex-zeros-of-frac-gammaszli-sz-pm-frac-gamma1-s
  [2]: http://en.wikipedia.org/wiki/Dirichlet_eta_function
  [3]: https://i.sstatic.net/YXJ5r.png
  [4]: http://en.wikipedia.org/wiki/Hurwitz_zeta_function