Extract imaginary part of $\text{Li}_3\left(\frac{2}{3}-i \frac{2\sqrt{2}}{3}\right)$ in closed form We know that polylogarithms of complex argument sometimes have simple real and imaginary parts, e.g.
$\mathrm{Re}[\text{Li}_2(i)]=-\frac{\pi^2}{48}$
Is there a closed form (free of polylogs and imaginary numbers) for the  imaginary part of
$\text{Li}_3\left(\frac{2}{3}-i \frac{2\sqrt{2}}{3}\right)$
 A: Inspired by this answer and by the comments below it, we could express it in terms of a generalized hypergeometric function as the following:
$$
\Im\left[\text{Li}_3\left(\frac{2}{3}-\frac{2\sqrt{2}}{3}i\right)\right] = \frac{1}{3}\arcsin^3\left(\frac{\sqrt3}{3}\right) - \frac{2\sqrt3}{3}{_4F_3}\!\left(\begin{array}c \tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32, \tfrac32,\tfrac32\end{array}\middle|\,\frac13\right).
$$
A: Inspired in turn by user 153012's answer which is similar to my answer in this post, then more generally, for any real $k>1$, 
$$\Im\left[\operatorname{Li}_3\left(\frac2k\,\big(1\pm\sqrt{1-k}\big)\right)\right] =\color{red}\mp\frac13\arcsin^3\left(\frac1{\sqrt k}\right)\pm\frac2{\sqrt k}\;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\frac1k\right)$$
where the OP's case was just $k=3$.
Edit: Courtesy of Oussama Boussif in his answer here, there is also a broad identity for $\rm{Li}_2(x)$ but for the real part,
$$\Re\left[\rm{Li}_{2}\left(\frac{1}{2}+iq\right)\right]=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{\left(\frac{1+4q^2}{4}\right)}}^{2}-\frac{{\arctan{(2q)}}^{2}}{2}
$$
