= Solution
If $M$ has complex dimension $n$, its complexified <cotangent bundle> splits into the $(1,0)$ and $(0,1)$ <cotangent bundles>. The space of global smooth <differential forms of type (p, q)> is
$$
\boxed{\mathcal A^{p,q}(M)=\Gamma_{C^\infty}\!\left(M,\Lambda^p(T^{1,0}M)^*\otimes\Lambda^q(T^{0,1}M)^*\right).}
$$
In <holomorphic coordinates>, its elements have the form $\alpha=\sum_{I,J}a_{I,J}\,dz^I\wedge d\bar z^J$, where the coefficients are complex-valued <smooth functions>, $|I|=p$, and $|J|=q$.
The <exterior derivative> decomposes as $d=\partial+\bar\partial$. The <conjugate Dolbeault operator> $\partial$ and <Dolbeault operator> $\bar\partial$ are its type components:
$$
\boxed{\partial\alpha=\sum_{I,J,k}\frac{\partial a_{I,J}}{\partial z_k}\,dz_k\wedge dz^I\wedge d\bar z^J,\qquad
\bar\partial\alpha=\sum_{I,J,k}\frac{\partial a_{I,J}}{\partial\bar z_k}\,d\bar z_k\wedge dz^I\wedge d\bar z^J.}
$$
They have bidegrees $(1,0)$ and $(0,1)$ respectively. Keeping the displayed wedge order avoids hiding a sign; moving $d\bar z_k$ past $dz^I$ introduces $(-1)^p$. Decomposing $d^2=0$ by type gives $\partial^2=\bar\partial^2=0$ and $\partial\bar\partial+\bar\partial\partial=0$.
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