# Wedge product $d(u\, dz)= \bar{\partial}u \wedge dz$.

How to show that if $u \in C_0^\infty(\mathbb{C})$ then

$d(u\, dz)= \bar{\partial}u \wedge dz$.

Note that $$d(u\,dz)=du\wedge dz+(-1)^0u\,ddz=du\wedge dz=(\partial u+\bar{\partial} u)\wedge dz.$$ Since $$\partial u=\frac{\partial u}{\partial z}dz\hspace{2mm}\mbox{ and }\hspace{2mm}\bar{\partial} u=\frac{\partial u}{\partial \bar{z}}d\bar{z},$$ we have $$d(u\,dz)=\frac{\partial u}{\partial z}dz\wedge dz+\frac{\partial u}{\partial \bar{z}}d\bar{z}\wedge dz=\frac{\partial u}{\partial \bar{z}}d\bar{z}\wedge dz=\bar{\partial} u\wedge dz,$$ where we have used $dz\wedge dz=0$.