Extract real and imaginary parts of $\operatorname{Li}_2\left(i\left(2\pm\sqrt3\right)\right)$

Solution 1:

The dilogarithms $\operatorname{Li}_2(z)$ and $\operatorname{Li}_2(1/z)$ are equal modulo elementary functions: $$\operatorname{Li}_2(z)+\operatorname{Li}_2(1/z)=-\frac{\pi^2}{6}-\frac{\ln^2(-z)}{2}.$$ Denoting $z_{\pm}=i(2\pm\sqrt3)$, we get in particular \begin{align}\operatorname{Li}_2\left(z_+\right)+\operatorname{Li}_2\left(-z_-\right) &=-\frac{\pi^2}{6}-\frac12\ln^2\left(z_-\right)=-\frac{\pi^2}{6}-\frac12\left(\ln |z_-|+\frac{i\pi}{2}\right)^2=\\ &=-\frac{\pi^2}{24}-\frac12\ln^2\left(2-\sqrt3\right)-\frac{i\pi}{2}\ln\left(2-\sqrt3\right).\tag{1} \end{align}

Taking the real and imaginary part of (1), one finds \begin{align} \Re\left(\operatorname{Li}_2\left(z_+\right)+\operatorname{Li}_2\left(z_-\right)\right)&=-\frac{\pi^2}{24}-\frac12\ln^2\left(2-\sqrt3\right). \tag{$\spadesuit$}\\ \Im\left(\operatorname{Li}_2\left(z_+\right)-\operatorname{Li}_2\left(z_-\right)\right)&=-\frac{\pi}{2}\ln\left(2-\sqrt3\right).\tag{$\clubsuit$} \end{align}

This already gives $66.67\%$ of Cleo's claim. Note that one could obtain similar relations for any purely imaginary $z_{\pm}$ satisfying $z_+z_-=-1$.

The remaining $33.33\%$ can be proved using another dilogarithm identity: $$\operatorname{Li}_2(z)-\operatorname{Li}_2(-z)+\operatorname{Li}_2\left(\frac{1-z}{1+z}\right)-\operatorname{Li}_2\left(-\frac{1-z}{1+z}\right)=\ln\frac{1+z}{1-z}\ln z+\frac{\pi^2}{4}.\tag{2}$$ The point is that if we set $z=z_-$ and take the imaginary part of (2), the first two dilogarithms give the contribution $2\Im\operatorname{Li}_2\left(z_-\right)$ whereas the arguments of the other two dilogarithms lie on the unit circle and are characterized by rational angles: $$2\Im\operatorname{Li}_2\left(z_-\right)+\color{blue}{\Im\left(\operatorname{Li}_2\left(e^{-i\pi/6}\right)-\operatorname{Li}_2\left(e^{5i\pi/6}\right)\right)}=\frac{\pi}{6}\ln\left(2-\sqrt3\right).\tag{3}$$ The piece outlined in blue can be found (this is precisely the place where the specific values of $z_{\pm}$ become important) using the triplication formula $$\frac13\operatorname{Li}_2\left(z^3\right)=\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(ze^{2\pi i/3}\right)+\operatorname{Li}_2\left(ze^{4\pi i/3}\right).$$ Namely, setting $z=e^{-i\pi/6}$, we get $$\frac13\operatorname{Li}_2\left(-i\right)=\operatorname{Li}_2\left(e^{-i\pi/6}\right)+\operatorname{Li}_2\left(i\right)+\operatorname{Li}_2\left(e^{-5i\pi /6}\right),$$ which in turn implies $$\Im\left(\operatorname{Li}_2\left(e^{-i\pi/6}\right)-\operatorname{Li}_2\left(e^{5i\pi/6}\right)\right)=-\frac43 \Im\operatorname{Li}_2\left(i\right)=-\frac{4}3\mathbf{G}.$$ Combining this with (3), we get $$\Im\operatorname{Li}_2\left(z_-\right)=\frac23\mathbf{G}+\frac{\pi}{12}\ln\left(2-\sqrt3\right)\tag{$\heartsuit$},$$ which finishes the proof. $\blacksquare$

Solution 2:

$$\Im\operatorname{Li}_2\left(i\left(2\pm\sqrt3\right)\right)=\frac{2\,{\bf G}}3-\frac{\pi\,(2\pm3)}{12}\ln\left(2-\sqrt3\right).$$ For the real part so far I could only establish the following identity: $$\Re\big[\operatorname{Li}_2\left(i\left(2+\sqrt3\right)\right)+\operatorname{Li}_2\left(i\left(2-\sqrt3\right)\right)\big]=-\frac{\pi^2}{24}-\frac{\ln^2\left(2-\sqrt3\right)}2.$$