Is this a holomorphic one-form:
$$A=\frac{1}{2i}B\bar{z}\hspace{1mm}dz?$$
where $B$ is a constant. MY ANSWER:
I read that a one-form $\omega=f(z,\bar{z})dz$ is holomorphic if $f(z,\bar{z})$ is holomorphic. Therefore, since in my example $\frac{1}{2i}\bar{z}$ does not satisfy the Cauchy-Riemann equations it is not holomorphic and therefore $A$ is not a holomorphic one-form.
I also found that a complex one-form $$\omega=f(z,\bar{z})dz$$ is holomorphic if $$\bar{\partial}\omega=0$$ where $$\bar{\partial}\omega=\frac{\partial f(z,\bar{z})}{\partial \bar{z}}dz\wedge d\bar{z}.$$ Therefore for my $A$, $$\bar{\partial}A=\frac{B}{2i}\frac{\partial \bar{z}}{\partial \bar{z}}dz\wedge d\bar{z}=\frac{B}{2i}dz\wedge d\bar{z}$$ therefore not holomorphic.
Any suggestions?