I'm following the Book of Brian Hatfield (Quantum Field Theory of point particles and Strings), p.192 here: For real Grassmann numbers (and Functionals thereof):
If $\Phi[\psi]$ is a functional, and $\psi(x)$ is a Grassmann-valued function, we demand that $$\int \mathcal{D}\psi \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]$$ (this is equation 9.67) , and one option to do this is to let (equation 9.66) $$\delta[\psi - \xi] = \prod_x (\psi(x)-\xi(x)).$$
The complex case of the delta function is NOT treated in the book, and I want to deduce how the mentioned relations would turn out for that case. Here, $\psi$ now has two components ($\psi = \frac{1}{2} \psi_1 + i \psi_2$) - Which makes me wonder: How does the fundamental relation turn out? For complex $\psi$: \begin{align} \int \mathcal{D}\psi \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi] \end{align} or \begin{align} \int \mathcal{D} \psi \int \mathcal{D} \psi^* \delta[\psi - \xi] \Phi[\psi] = \Phi[\xi]? \end{align}
The first version works, but only if I assume that $\delta[\psi - \xi] = \prod\limits_x (\psi(x) - \xi(x))$ and $\delta[\psi-\xi] = \delta[\psi-\xi]^*$, and those exclude each other. In either case, what would be a realization of the $\delta$ functional? Would it still be \begin{align} \delta[\psi - \xi] = \prod_x (\psi(x) - \xi(x))? \end{align}