Let $\mathbb C^2$ denote a two-dimensional complex Euclidean space with a Hermitian metric $h=\sum\limits_{i,j=1}^{2}h_{i\bar j}dz^i\otimes d\bar z^j$, where $h_{i\bar j}$ are constant numbers, then the associated fundamental form is $\omega=\frac{\sqrt{-1}}{2}\sum\limits_{i,j=1}^{2}h_{i\bar j}dz^i\wedge d\bar z^j$. Let $\omega_1,\omega_2,\omega_3,\omega_4$ be four vectors in $\mathbb C^2$ which are linearly independent over $\mathbb R$, where $\omega_i=(\omega_i^1,\omega_i^2)\in \mathbb C^2$. Let $\Lambda$ be the lattice generated by $\omega_1,\omega_2,\omega_3,\omega_4$, then we can get a complex torus $T=\mathbb C^2/\Lambda$. Any two vectors of these four vectors determine a parallelogram in $\mathbb C^2$, for example, if we choose $\omega_1,\omega_2$, let $D_{12}$ denote the parallelogram spanned by $\omega_1,\omega_2$, then we want to compute the integral of the fundamental form $\omega$ over $D_{12}$, i.e. $\int_{D_{12}}\omega$.
According to Chern's book Complex manifolds without potential theory. 2nd edition. p.65 Formula 7.30, the integral may be equal to $$g_{12}=i\sum_{j,k}h_{j\bar k}(\omega^j_1\bar\omega^k_2-\omega^j_2\bar\omega^k_1)$$ (I doubt $i$ should be replaced by $\frac{i}{2}$, see the following). And according to Wells' book Differential analysis on complex manifolds. Third edition p.221. the period of $\omega$ over the 2-cycle is given by $$\int_{D_{12}}\omega=\frac{i}{2}\sum_{j,k}h_{j\bar k}(\omega^j_1\bar\omega^k_2-\omega^j_2\bar\omega^k_1).$$
Here my question is: why the integral $\omega$ over the parallelogram $D_{12}$ takes such form? And does it relate to $\omega(u,v),u,v\in T_{\mathbb C^2}$?