$\text{tr}\log(A)+\text{tr}\log(B)$ for complex matrices? For positive definite matrices $A$ and $B$, we have the following identity
$$\text{tr}\log(A)+\text{tr}\log(B)=\text{tr}\log(AB).$$
What can be said about the case for complex matrices $A$ and $B$? Assume they are still invertible.
 A: If matrix logarithm is defined as a primary matrix function and the principal branch of logarithm is selected, the trace identity is still valid when all eigenvalues of $A,B$ and $AB$ are positive (we don't need $A$ or $B$ to be positive definite). This is because $\operatorname{tr}\log(M)=\log(\det(M))$ when $M$ has a positive spectrum.
If matrix logarithm is defined as a primary matrix function but at least one of $A,B$ or $AB$ has a non-positive eigenvalue, it can happen that the trace identity always fails regardless of the branch of logarithm selected. For example, let $\omega=\exp(2\pi i/3),\,A=\operatorname{diag}(1,\omega,\omega^2)$ and $B=\omega I_3$. Then $AB=\operatorname{diag}(\omega,\omega^2,1)$ and
\begin{aligned}
\log(A)&=\operatorname{diag}\left(\log(1),\log(\omega),\log(\omega^2)\right),\\
\log(B)&=\operatorname{diag}\left(\log(\omega),\log(\omega),\log(\omega)\right),\\
\log(AB)&=\operatorname{diag}\left(\log(\omega),\log(\omega^2),\log(1)\right).
\end{aligned}
Hence
\begin{aligned}
\operatorname{tr}\log(AB)
&=\log(1)+\log(\omega)+\log(\omega^2)\\
&\ne\log(1)+4\log(\omega)+\log(\omega^2)\\
&=\operatorname{tr}\log(A)+\operatorname{tr}\log(B),
\end{aligned}
because $\log(\omega)$ is nonzero for all branches of logarithm.
