What is the cohomology of the $C^{\infty}$-Koszul complex on $\mathbb{C}^2$? First we consider the holomorphic Koszul complex on $\mathbb{C}^2$:
$$
0\to \mathcal{O}(\mathbb{C}^2)\overset{\begin{pmatrix}-z_2\\z_1\end{pmatrix}}{\to} \mathcal{O}(\mathbb{C}^2)^{\oplus 2}\overset{(z_1,z_2)}{\to} \mathcal{O}(\mathbb{C}^2)\to \mathbb{C}\to 0.
$$
It is well-known that the above complex is exact.
Now we consider the $C^{\infty}$-version of the Koszul complex
$$
0\to C^{\infty}(\mathbb{C}^2)\overset{\begin{pmatrix}-z_2\\z_1\end{pmatrix}}{\to} C^{\infty}(\mathbb{C}^2)^{\oplus 2}\overset{(z_1,z_2)}{\to} C^{\infty}(\mathbb{C}^2)\to \mathbb{C}\to 0.
$$
It is clear that the above complex is not exact. For example $\bar{z_1}$ and $\bar{z_2}$ both vanish at the origin but are not in the image of multiplying $z_1$ and $z_2$.


My question is: What is the cohomology of the above complex?


 A: As Vladimir mentions in the comment, it is probably easier to formulate this without the $\mathbb{C}$ at the right. However, it is not clear to me that the answer then is exactly what Vladimir claims, i.e., that you get the antiholomorphic functions as cohomology at level $0$.
I am not sure if you intend the statement to be for global sections, as I interpret your notation, but since the involved sheaves $\mathbb{C}$ and $C^\infty$ are acyclic on $\mathbb{C}^2$, and the complexes are exact outside the origin, you might just as well consider the corresponding complexes of stalks at the origin.
By flatness of the smooth functions over the holomorphic functions, it follows that you have an exact complex of sheaves
$$0 \to C^\infty_0 \to (C^\infty_0)^{\oplus 2} \to C^\infty_0 \to C^\infty_0/((z_1,z_2)C^\infty_0) \to 0,$$
where $C^\infty_0$ denote the ring of germs of smooth functions at $0$.
The map from the antiholomorphic functions, $\overline{\mathcal{O}_0} \to C^\infty_0/((z_1,z_2)C^\infty_0)$ is injective, since $\overline{\mathcal{O}_0} \cap (z_1,z_2)C^\infty_0 = \\{0\\}$ (which you might verify by reducing to the one-dimensional case, by restricting to complex lines to the origin). However, it is not so clear to me that the map is actually surjective.
