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If two numbers $\eta$ and $\xi$ anti-commute. i.e., $$\eta\xi=-\xi\eta$$ they are called Grassmann numbers. It immediately follows that $$\eta^2=\xi^2=0,$$ and relations such as $$e^{a\eta}=1+a\eta;~~e^{a\eta\xi}=1+a\eta\xi~~\text{etc}$$ and a whole lot of interesting properties.

One often talks about complex Grassmann numbers $\eta$, in higher dimensions. For example, in two dimensions, one defines as $\eta=\begin{pmatrix}\eta_1\\ \eta_2\end{pmatrix}$ and $\xi=\begin{pmatrix}\xi_1\\ \xi_2\end{pmatrix}$. In this case, one can define products such as $\eta^T\xi$ and $\eta^\dagger\xi$ which gives $$\eta^T\zeta=\eta_1\xi_1+\eta_2\xi_2;~~\eta^\dagger\zeta=\eta^*_1\xi_1+\eta_2^*\xi_2.$$

In the first case, it is quite clear that since $\eta_1^2=\eta_2^2=0$, the relation $\eta^T\eta$ (sometimes abbreviated as $\eta\eta$) gives zero. However, I've trouble with evaluating the product $\eta^\dagger\eta=\eta^*_1\eta_1+\eta_2^*\eta_2$ where $\eta_{1}$ ($\eta_2$) and $\eta_1^*$ ($\eta_2^*$) are taken to be two independent Grassmann numbers so that $$\eta_1^*\eta_1=-\eta_1\eta_1^*.$$

My question is whether $\eta^\dagger\eta$ reduce to zero. The way I did it, it's not. But I'm not sure whether I'm making a mistake. Please help me with this.

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It seems OP is effectively asking the following.

Given a complex Grassmann-odd variable $z=x+iy$ and its complex conjugate variable $z^{\ast}=x-iy$, is the product $z^{\ast} z$ zero?

Answer: Not necessarily:

$$ z^{\ast} z~=~(x-iy)(x+iy)~=~\ldots~=~2ixy. $$

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