Let $M$ be a holomorphic (complex) manifold with: $$(1)\quad dim_{C}(M)=m. $$
What I understand regarding a holomorphic form is that it is:
$$(2)\quad \alpha^{(r,0)}=\frac{1}{r!}f_{\mu_1,\dots,\mu_r}(z)\space dz^{\mu_1}\wedge\dots\wedge dz^{\mu_r}\in\Omega^{(r,0)}(M). $$
Here the totally antisymmetric component functions are themselves holomorphic, and consequently entail only $z$ dependence.
Similarly an antiholomorphic form is given by:
$$(3)\quad \beta^{(0,s)}=\frac{1}{s!}g_{\bar{\nu}_1,\dots,\bar{\nu}_s}(\bar{z})\space d\bar{z}^{\nu_1}\wedge\dots\wedge d\bar{z}^{\nu_s}\in\Omega^{(0,s)}(M), $$ of course with the component functions being totally antisymmetric and antiholomorphic, consequently entailing only $\bar{z}$ dependence.
Taking the wedge of $(2)$ and $(3)$, we get the generic $(r,s)$ form: $$(4)\quad \omega^{(r,s)}=\frac{1}{r!s!}f_{\mu_1,\dots,\mu_r}(z)g_{\bar{\nu}_1,\dots,\bar{\nu}_s}(\bar{z})\space dz^{\mu_1}\wedge\dots\wedge dz^{\mu_r}\wedge d\bar{z}^{\nu_1}\wedge\dots\wedge d\bar{z}^{\nu_s}\in\Omega^{(r,s)}(M). $$
Now a form is holomorphic if and only if $\bar{\partial}\omega=0$.
If I begin with the form in $(2)$, which is holomorphic, then there is no problem seeing that $\bar{\partial}\alpha^{(r,0)}=0$, since this would involve $\bar{z}$ derivatives of the component functions which are clearly vanishing.
The problem arises when I begin with $\eta^{(p,q)}\in\Omega^{(p,q)}(M)$, require that it satisfy $\bar{\partial}\eta^{(p,q)}=0$, and try to get a quantity of the form $(2)$.
Question 1: Can someone please guide me along this proof?
Question 2: The following is particularly bothering me: consider $\gamma^{(p,m)}\in \Omega^{(p,m)}(M)$, then clearly $\bar{\partial}\gamma^{(p,m)}=0$ (for the wedge having $m+1$ barred indices is vanishing), but I cannot write it in the form $(2)$.