Let be $f:U\to \Bbb C$ antiholomorphic function. Show that the 1-form $f(z)d\overline{z}$ is closed.
We have that $\overline{f}$ is a holomorphic function, so by Cauchy-Goursat Theorem the 1-form $\overline{f}(z)dz$ is closed. But, how can I conclude that $f(z)d\overline{z}$ is closed?
Thanks!
Differentiation is a real operator, that is, it commutes with conjugation,
$$d\overline{\omega} = \overline{d\omega}$$
for every $k$-form. Since you know that $\overline{f(z)}\,dz$ is closed, i.e. $d\bigl(\overline{f(z)}\,dz\bigr) = 0$, the result follows.
Alternatively, use the Wirtinger derivatives,
$$dg = \frac{\partial g}{\partial z}\,dz + \frac{\partial g}{\partial \overline{z}}\,d\overline{z},$$
and note that $f$ being antiholomorphic means $\frac{\partial f}{\partial z} = 0$, so
$$d\bigl(f(z)\,d\overline{z}\bigr) = \bigl(df(z)\bigr)\wedge d\overline{z} = \biggl(\frac{\partial f}{\partial z}(z)\,dz + \frac{\partial f}{\partial \overline{z}}(z)\,d\overline{z}\biggr)\wedge d\overline{z} = 0.$$
