# Li 2(z)=-Li 2(1/z)-(1/2)(log z)^2 + i pi log(z) + pi^2/3

The following formula holds: $$\mathrm{Li}_2(z)=-\mathrm{Li}_2 \left( \dfrac{1}{z} \right) - \dfrac{\log(z)^2}{2} + i \pi \log(z) + \dfrac{\pi^2}{3},$$ where $\mathrm{Li}_2$ denotes the dilogarithm, $\log$ denotes the logarithm, $i$ denotes the imaginary number, and $\pi$ denotes pi.