Basis for the alternating vector space $\mathcal{B}(\mathbb{R}^2\times\mathbb{R}^2,\mathbb{C})$

Question

I am currently reading Complex variables an introduction by Bernstein, and Carlos.A.

If $ B \in \mathcal{B}(\mathbb{R}^2\times\mathbb{R}^2,\mathbb{C})$. Then we have I calculated $(B(\vec{e}_1,\vec{e}_2)dx \wedge dy)(h_1,h_2) = B(\vec{e}_1,\vec{e}_2)h_1k_2 - B(\vec{e}_1,\vec{e}_2)(k_1,h_2)$

Why is that equal to $B(h_1,h_2)$? Recall $dx(w) = Re(w)$ and $dy(w) = im(w)$ for $w \in \mathbb{C}$.

Answer 1

I am sorry it is actually very simple.

$B(\vec{e}_1,\vec{e}_2)h_1k_2 - B(\vec{e}_1,\vec{e}_2)k_1h_2 = B(\vec{e}_1h_1,\vec{e}_2k_2) - B(\vec{e}_1k_1,\vec{e}_2h_2) = B(\vec{e}_1h_1,\vec{e}_2k_2) + B(\vec{e}_2h_2,\vec{e}_1k_1) = B(h,k)$