Complex gaussian integral with a complex action and different source terms

Question

I am trying to use the following Gaussian path integral identity $$\int D[\phi_1,\phi_1^*,\cdots,\phi_n,\phi_n^*] \exp(i\int z^\dagger D z+i\int f^\dagger z+z^\dagger g) = \det{D}^{-1}\exp(-i\int f^\dagger D^{-1}g).$$ This works if $z$ is a vector of complex fields $z=(\phi_1,\cdots,\phi_n)^T$. However, in my case I need z to contain both the fields and their complex conjugates, i.e., $z=(\phi_1,\phi_1^*,\cdots, \phi_n,\phi_n^*)^T$. In such a case, I dont know how to distribute the source terms between the $f^\dagger$ and $g$ terms, i.e., I could easily choose $f^\dagger=0$ or $g=0$. Hence I think this identity clearly does not apply to this case. In other words, the integral I have is $$\int D[\phi_1,\phi_1^*,\cdots,\phi_n,\phi_n^*] \exp(i\int z^\dagger D z+i\int f^\dagger z)$$ where $z=(\phi_1,\phi_1^*,\cdots, \phi_n,\phi_n^*)^T$ but $f^\dagger \neq (f_1,f_1^*,\cdots)^T$. Another way of stating this is that the Euler-Lagrange equations are not complex conjugate of each other.