For a smooth function f , one form $df$ is defined as $df(Z)= fd(Z)$.
Prove that:
$$ df= \frac{ \partial f}{\partial z}dz+ \frac{\partial f} {\partial \bar z} d\bar z$$
My attempt:
One-form $df$ can be expressed in terms of basis as:
$$df=adz+ bd \bar z$$ $$Z(f)= (\frac{ \partial f}{\partial z}, \frac{\partial f} {\partial \bar z})$$ $$df(Z)=(a,b)$$ Thus, $$ df= \frac{ \partial f}{\partial z}dz+ \frac{\partial f} {\partial \bar z} d\bar z$$
I am really confused, if my attempt is correct. Any help will be very appreciated. Thank you!!