How do I solve this Gaussian path integral? Suppose
$$
Z = \int \mathcal D[\phi^*] \mathcal D[\phi] \exp(\phi^*A\phi + \phi B\phi)
$$
where $A$ and $B$ are operators. I know how to solve a Gaussian path integral involving only $\phi^* A \phi$ but I don't know how to handle the other quadratic term.
 A: You just divide $\phi$ to the real and imaginary part, to make things clear:
$$\phi =  f + ig, \quad \phi^* = f-ig, \quad f,g\in{\mathbb R}$$
Up to some totally universal normalization factor, the integration measure is simply
$$\int {\mathcal D} f \,\,{\mathcal D} g $$
and the exponent in the exponential may be written as
$$[(f-ig) A + (f+ig) B] (f+ig) $$ 
Writing the column $(f,g)^T$ as $h$, the bilinear expression above is nothing else than
$$ h M h $$
where the matrix $M$ is, in a block-diagonal form,
$$ M = \left(\begin{array}{cc}A+B&-iA+iB\\iA+iB&A-B\end{array}\right) $$
Now, I assume you may calculate the integral 
$$\int {\mathcal D} h\,\exp(hMh) $$
which is completely analogous to the $\exp(\phi^* A \phi)$ integral. However, with the matrix $M$ enough, the integral is infinity because $M$ is singular (infinity, due to flat directions) because the second row (of blocks) is $i$ times the first. However, you will get a nonsingular result if the exponent will also contain the Hermitian conjugate $\phi^* B^\dagger \phi^*$ or something like that.
If there are algebraic mistakes above, it should be possible to fix them.
