Proving that a (complex) differential form is of type $(p, q)$ iff its conjugate is of type $(q, p)$ using complex vector fields Let $(M, J)$ be a $n$-complex manifold and $p+q=k$ where $0\leq k \leq 2n$. I'm looking for a clean way to prove that $\mu\in \mathcal{A}^k(M, \mathbb{C}):=\Gamma\left(\bigwedge^k T_{\mathbb C}^*M \right)$ is a complex diferrential $k$-form of type $(p, q)$ iff $\bar{\mu}$ is of type $(q, p)$. Now, first of all, I want to do everything avoiding real arguments in a coordinate-free fashion using complex vector fields (i.e, smooth sections of the complexified tangent bundle of $M$,  which I will denote by $\mathfrak{X}(M, \mathbb C):=\Gamma\left(T_{\mathbb C}M \right)$).
Ok, so firstly, we define $\bar{\mu}\in \mathcal{A}^k(M, \mathbb{C})$ by
$$(1)\qquad\qquad\qquad
 \bar{\mu}(Z_1,\ldots,Z_k):= \overline{\mu(\bar{Z_1},\ldots,\bar{Z_k})},\qquad\qquad  Z_1,\ldots,Z_k\in \mathfrak{X}(M, \mathbb C).$$
I'm aware that one can check that a $k$-form is a $(p,q)$-form when $\lambda\star \mu= \lambda^p \bar{\lambda}^q \mu$, with $\lambda \in \mathbb C$ (or more generally, for $\lambda \in \mathcal{C}^\infty(M, \mathbb C)$) and where $\lambda\star \mu\in \mathcal{A}^k(M, \mathbb{C}) $ is defined as
$$(2)\qquad\qquad\qquad\lambda\star \mu (Z_1,\ldots,Z_k):= \mu(\lambda Z_1,\ldots, \lambda Z_k).$$
Now, the trouble comes here. If we want to check (2) applied to (1), we end up obtaining that $\lambda\star \bar\mu=\lambda^p\bar{\lambda}^q \bar\mu$, which is definitely NOT what we want. However, if we instead define $\bar \mu$ by
$$(3)\qquad\qquad\qquad
 \bar{\mu}(Z_1,\ldots,Z_k):= \overline{\mu(Z_1,\ldots,Z_k)},$$
then we arrive at the desired conclusion.
Now, in fact I've taken definition $(3)$ for granted for some time, but now I'm convinced it is NOT the right one.
If we play with a toy example, let's say $ \mu = dz\wedge d\bar{z}$ and $Z=a\dfrac{\partial}{\partial z}+b\dfrac{\partial}{\partial \bar{z}}, W=\alpha\dfrac{\partial}{\partial z}+\beta\dfrac{\partial}{\partial \bar{z}}$, we all agree that $\bar\mu= d\bar{z}\wedge dz$, and we can easily check that $\bar\mu(Z, W)=\overline{\mu(\bar Z, \bar W)}=b\alpha-a\beta$, whereas $\overline{\mu(Z, W)}=\overline{a\beta - b \alpha}$, so definition $(1)$ is the right one.
So, what am I missing here? I'm certain that the conjugate is given by $(1)$. Not many authors bother to explain what $\bar\mu$ actually is, or how to compute it, a small portion of them just say that if $\mu=\alpha+i \beta$, then $\bar\mu=\alpha -i\beta$ (which I agree with, but it is quite dry). In fact, formula $(1)$ appears in Poor's Differential Geometric Structures, a quite venerable and underrated text, and as we saw in the toy example, it is quite the right one.
All that being said, we can infer that maybe $(2)$ doesn't work as expected for complex vector fields. If so, what can we do so that $(2)$ works for my desired proof?
I guess we can just take the following definition for a $(p, q)$-form and everything is settled:
A complex $k$-form is of type $(p, q)$ if and only if it vanishes whenever applied
to $p+1$ vectors of type $(1, 0)$ or to $q + 1$ vectors of type $(0, 1)$.
The desired conclusion holds trivially using the previous definition, but I'm quite unsatisfied because $(2)$ is a pretty nice way to check if a $k$-form is a $(p, q)$-form. Any help or comments are appreciated. Just to be clear, I don't have trouble with the fact that $\mu$ is a $(p, q)$-form iff $\bar \mu$ is of type $(q, p)$   (for example, using the definition above or local coordinates), rather, I want to see why $(2)$ doesn't hold or what does it need to work correctly in the fashion I want to use it.
 A: This is basically a question on linear algebra. Let $V$ be a finite dimensional real vector space and let $\mu \in \Lambda^k(V_{\mathbb{C}}^{*})$. There are two ways to think about $\mu$:

*

*The first is as an alternating real multi-linear map $\mu \colon V \times \dots \times V \rightarrow \mathbb{C}$. Note that the input of $\mu$ are "real" vectors in $V$ and the output is a complex number. This corresponds to thinking of $\mu$ as an element of $\left( \Lambda^k_{\mathbb{R}}(V^{*}) \right)_{\mathbb{C}} = \Lambda^k_{\mathbb{R}}(V^{*}) \otimes \mathbb{C}$.

*The second is an alternating complex multi-linear map $\hat{\mu} \colon V_{\mathbb{C}} \times \dots \times V_{\mathbb{C}} \rightarrow \mathbb{C}$. Here the inputs are complexified vectors and the output is a complex number. The complex structure of $V_{\mathbb{C}}$ comes from the complexification (and not from $J$!). This corresponds to thinking of $\mu$ as an element of $\Lambda^k_{\mathbb{C}}(V_{\mathbb{C}}^{*})$.

Both ways of thinking about $\mu$ are equivalent and useful. The relation between $\mu$ and $\hat{\mu}$ is given by "extending $\mathbb{C}$-linearly". Namely, for example, for $k = 1$,
$$ \hat{\mu} \left( v + iw \right) = \mu(v) + i\mu(w). \label{eq:single} $$

So let's think about $\mu$ as in $(1)$. The vector space $V$ has no natural conjugation but $\mathbb{C}$ does so we can define
$$ (1) \qquad\qquad \overline{\mu} \left( v_1, \dots, v_k \right) = \overline{ \mu \left( v_1, \dots, v_k \right)}. $$
This is another real multi-linear map so everything is well-defined.
Now let's think about $\hat{\mu}$ as in $(2)$. In this case, both $V_{\mathbb{C}}$ and $\mathbb{C}$ carry natural conjugations and if we want $\overline{\hat{\mu}}$ to be $\mathbb{C}$-linear, we are forced to use them both and define
$$ (2) \qquad\qquad \overline{\hat{\mu}} \left( v_1, \dots, v_k \right) = \overline{ \hat{\mu} \left( \overline{v_1}, \dots, \overline{v_k} \right)}. $$
This is indeed a $\mathbb{C}$ multi-linear map. Under the isomorphism $\mu \mapsto \hat{\mu}$, the "single conjugation" $\overline{\mu}$ corresponds to the "double conjugation" $\overline{\hat{\mu}}$.
The bottom line is that whenever you work with real vectors, you use $(1)$ and whenever you work with complexified vectors, you use $(2)$. If the vectors in $(2)$ are real, then clearly the inner conjugation does nothing.

Finally, let's introduce a complex structure $J$ on $V$. It seems you have misunderstood the $(p,q)$ criteria. In order to check whether $\mu$ (thought of as in $(1)$ above) is a $(p,q)$ form, you define for $\lambda = a + ib \in \mathbb{C}$
$$ (3) \qquad\qquad \lambda \star \mu (v_1,\ldots,v_k):= \mu \left( av_1 + bJv_1,\ldots, av_k + bJv_k \right) $$
and then check whether $\lambda \star \mu = \lambda^p \overline{\lambda}^q \mu$.
In order to check whether $\hat{\mu}$ (thought of as $(2)$ above) is a $(p,q)$ form, you define
$$ (4) \qquad\qquad \lambda \star \hat{\mu} (v_1,\ldots,v_k):= \mu \left( av_1 + b\hat{J}v_1,\ldots, av_k + b\hat{J}v_k \right) $$
where $v_1,\dots,v_k$ are complexified vectors and $\hat{J}$ is the $\mathbb{C}$-linear extension of $J$ to $V_{\mathbb{C}}$ and then check whether $\lambda \star \mu = \lambda^p \overline{\lambda}^q \mu$. Note that this is not the same as
$$ \hat{\mu} \left( \lambda v_1, \dots, \lambda v_k \right) = \lambda^k \hat{\mu} \left( v_1, \dots, v_k \right) $$
(as $\hat{\mu}$ is $\mathbb{C}$-linear).
Finally, if you use $(2)$ and $(4)$, you see that if $\hat{\mu}$ is a $(p,q)$ form then
$$ \left( \lambda \star \overline{\hat{\mu}} \right) \left( v_1, \dots, v_k \right) = \overline{ \hat{\mu} \left( \overline{av_1 + b\hat{J}v_1}, \dots, \overline{av_k + b\hat{J}v_k} \right)} = 
\overline{ \hat{\mu} \left( a\overline{v_1} + b\hat{J}\overline{v_1}, \dots, a\overline{v_k} + b\hat{J}\overline{v_k} \right)} = \overline{ \lambda^p \overline{\lambda}^q} \overline{\hat{\mu}} (v_1, \dots, v_k) = \lambda^q \overline{\lambda}^p \overline{\hat{\mu}} (v_1, \dots, v_k)$$
and so indeed $\hat{\mu}$ is a $(q,p)$-form.
