Question on specific properties of inner product on complex manifolds

Question

I am having trouble understanding part of a proof within Principles of Algebraic Geometry; GRIFFITHS / HARRIS. The proof I am struggling with is located at page 112 under the subitem The Hodge Identities and the Hodge Decomposition.

There, the operator

$$e_k : A^{p,q}_c( \mathbb{C}^n) \longrightarrow A^{p+1,q}_c (\mathbb{C}^n) \; , \; \; e_k(\varphi) = dz_k \wedge \varphi$$

and and the respective complex conjugate $\bar{e}_k$ and adjoint $i_k$ are introduced. Furthermore, the following two identities are stated whose reasoning I do not understand:

1. $$i_k( dz_J \wedge d\bar{z}_K ) \enspace = \enspace 0$$
2. $$i_k( dz_k \wedge dz_J \wedge d\bar{z}_K ) \enspace = \enspace 2 dz_J \wedge d\bar{z}_K$$

Now, these identities are justified by saying that for the inner product it holds that

1. $$ \Big( i_k( dz_J \wedge d\bar{z}_K ), dz_L \wedge d\bar{z}_M \Big) \enspace = \enspace \Big( dz_J \wedge d\bar{z}_K, dz_k \wedge dz_L \wedge dz_M \Big) \enspace = \enspace 0$$
2. $$ \Big( i_k( dz_k \wedge dz_J \wedge d\bar{z}_K ), dz_L \wedge d\bar{z}_M \Big) \enspace = \enspace \Big( dz_k \wedge dz_J \wedge d\bar{z}_K , dz_k \wedge dz_L \wedge d\bar{z}_M \Big) \enspace = \enspace 2 \Big( dz_J \wedge d\bar{z}_K, dz_L \wedge d\bar{z}_M \Big)$$

My questions are:

i) Why does $\Big( dz_J \wedge d\bar{z}_K, dz_k \wedge dz_L \wedge dz_M \Big)$ vanish?

ii) Why does $\Big( dz_k \wedge dz_J \wedge d\bar{z}_K , dz_k \wedge dz_L \wedge d\bar{z}_M \Big) = 2 \Big( dz_J \wedge d\bar{z}_K, dz_L \wedge d\bar{z}_M \Big)$ hold?

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These maybe some easy questions for those who really understand the subject. However, I am still trying to get "accustomed" to working on complex manifolds and its implications. I therefore am a bit lost here.