Since you are looking for the imaginary part of the equation, the most important hint is that the polylogarithm $\operatorname{Li}_{s}(z)$of order $s \in \mathbb{R}$ is real if $ z< 1$ and complex if $z>1$ for $z\in \mathbb{R}$.
And the values of the imaginary part can be obtained from the formula:
$$ \boxed{\Im \operatorname{Li}_{s}(z) = -\frac{\pi \ln^{s-1}(x)}{\Gamma(s)}, \quad z>1}$$
Lets prove the case $s=2$. By definition:
$$\operatorname{Li}_2(z) = -\int_{0}^{z}\frac{\ln(1-x)}{x}dx$$
Suppose that $z>1$
$$\operatorname{Li}_2(z) = -\int_{0}^{1}\frac{\ln(1-x)}{x}dx - \int_{1}^{z}\frac{\ln(1-x)}{x}dx$$
However $\displaystyle \ln(1-x)= \ln|1-x| + i\pi$ for $x>1$ where we take the principal branch of $\displaystyle \mathbf{\ln}$:
\begin{eqnarray*}
\operatorname{Li}_2(z) &=& -\int_{0}^{1}\frac{\ln(1-x)}{x}dx - \int_{1}^{z}\frac{\ln|1-x|}{x}dx - i\pi\int_{1}^{z}\frac{1}{x}dx \\
&=&-\int_{0}^{1}\frac{\ln(1-x)}{x}dx - \int_{1}^{z}\frac{\ln|1-x|}{x}dx - i\pi\ln(z)
\end{eqnarray*}
Taking the imaginary part in both sides:
$$\boxed{ \Im \operatorname{Li}_{2}(z) = -\pi\ln(z), \quad z>1} $$
Examples:
$$ \Im \operatorname{Li}_{3}(3) = -\frac{\pi}{2}\ln^2(3)$$
$$ \Im \operatorname{Li}_{2}\left(\frac{3}{2}\right) = -\pi\ln\left(\frac{3}{2}\right) = -\pi\ln(3)+\pi\ln(2) $$
