On a property of polylogarithm

$\def\Li{{\rm Li}}$Note that generally $\Li_n$ is single valued analytic function in $D=\mathbb{C}\setminus[1,+\infty)$. This domain is symmetric with respect to the real axis. Now, $g(z)=\overline{\Li_n(\overline{z})}$ is also analytic in the same domain $D$ and it coincides with $\Li_n$ on the real interval $(-1,1)$. This implies that $g=\Li_n$ on $D$. $$ \forall\,z\in D,\qquad \Li_n(z)=\overline{\Li_n(\overline{z})} $$ From this we conclude that $$\eqalign{ \Re(\Li_n(z))&=\frac{1}{2}\left(\Li_n(z)+\overline{\Li_n(z)}\right)= \frac{1}{2}\left(\Li_n(z)+\Li_n(\overline{z})\right)\cr \Im(\Li_n(z))&=\frac{1}{2i}\left(\Li_n(z)-\overline{\Li_n(z)}\right)= \frac{1}{2i}\left(\Li_n(z)-\Li_n(\overline{z})\right)\cr } $$ In particular, for $z\in D$, we have $$\eqalign{ \Li_n(z)+\Li_n(\overline{z})&\in\mathbb{R}\cr \Li_n(z)-\Li_n(\overline{z})&\in \mathbb{R} i } $$ from which the desired conclusion follows.