Denote $\operatorname{Li}_4(z)$ the analytic continuation of $\sum_{n=1}^\infty\frac{z^n}{n^4}$. $z$ is a algebraic number with $|z|\ne 0,1$. Does $\Re\operatorname{Li}_4(z)$ or $\Im\operatorname{Li}_4(z)$ have closed-form with some $z$?
The following is something I've found. $$\operatorname{Li}_4(0)=0$$ $$\operatorname{Li}_4(1)=\zeta(4)$$ $$\operatorname{Li}_4(-1)=-\eta(4)$$ $$\operatorname{Li}_4(i)=-\frac{7 \pi ^4}{11520}+\frac{i \psi ^{(3)}\left(\frac{1}{4}\right)}{1536}-\frac{i \psi ^{(3)}\left(\frac{3}{4}\right)}{1536}$$ $$\operatorname{Li}_4(-i)=-\frac{7 \pi ^4}{11520}-\frac{i \psi ^{(3)}\left(\frac{1}{4}\right)}{1536}+\frac{i \psi ^{(3)}\left(\frac{3}{4}\right)}{1536}$$ $$\Re\operatorname{Li}_4(e^{ix})=-\frac{x^4}{48}+\frac{\pi x^3}{12}-\frac{\pi ^2 x^2}{12}+\frac{\pi ^4}{90}$$ But I failed to give another non-trivial examples. I thought about $\Re\operatorname{Li}_4\left(\frac{1+i}2\right)$ and use the same method evaluating $\int_0^1 \frac{x \log ^2(x+1)}{x^2+1} \, dx$ ($x\mapsto\frac{1-x}{1+x}$) to evaluate $$\int_0^1 \frac{x \log ^3(x+1)}{x^2+1} \, dx$$ but failed.
For $z=0, e^{2\pi i \phi}(\phi\in\mathbb{Q})$ the tetralog have closed form in terms of polygamma functions. For $z=1\pm i, 1/2\pm i/2, e^{i \varphi}(\varphi\in\mathbb{R})$ its real part have closed form (see for instance here): $$\Re \text{Li}_4(1\pm i)=-\frac5{16} \text{Li}_4\left(\frac{1}{2}\right)+\frac{97 \pi ^4}{9216}-\frac{5}{384} \log^4(2)+\frac{1}{48} \pi ^2 \log^2(2)$$ $$\Re \text{Li}_4 \left (\frac{1 \pm i}{2} \right ) =\frac{5}{16} \text{Li}_4 \left (\frac{1}{2} \right ) + \frac{343 \pi^4}{92160} + \frac{1}{96} \log^4 (2)- \frac{5}{768} \pi^2\log^2 (2)$$