[proofplan] We compute both sides on the real coordinate frame associated to the holomorphic chart. First we express the real metric coefficients of $g$ in terms of the Hermitian coefficients $h_{i\bar j}$ using the convention $g=2\operatorname{Re}h$. Then the identities $J\partial_{x_i}=\partial_{y_i}$ and $J\partial_{y_i}=-\partial_{x_i}$ give the real coordinate coefficients of $\omega$. Finally we expand the proposed complex-coordinate expression and verify that it has exactly the same values on the real frame. [/proofplan]

[step:Express the real coordinate frame in terms of the holomorphic frame] Fix the coordinate chart $(U,z)$ and write $z_i=x_i+i y_i$. For each $i\in\{1,\dots,n\}$, define the real coordinate vector fields \begin{align*} X_i=\partial_{x_i} \end{align*} and \begin{align*} Y_i=\partial_{y_i}. \end{align*} The associated holomorphic coordinate vector field is \begin{align*} \partial_{z_i}=\frac{1}{2}(X_i-iY_i). \end{align*} The $(1,0)$-parts of the real vector fields are therefore \begin{align*} (X_i)^{1,0}=\partial_{z_i} \end{align*} and \begin{align*} (Y_i)^{1,0}=i\partial_{z_i}. \end{align*} The complex structure acts on the real coordinate frame by \begin{align*} JX_i=Y_i \end{align*} and \begin{align*} JY_i=-X_i. \end{align*} [/step]

[step:Compute the real metric coefficients from the Hermitian coefficients]For $i,j\in\{1,\dots,n\}$, set \begin{align*} h_{i\bar j}=h(\partial_{z_i},\partial_{z_j}). \end{align*} Since $h$ is Hermitian with the stated convention, $h_{j\bar i}=\overline{h_{i\bar j}}$. Using $g(u,v)=2\operatorname{Re}h(u^{1,0},v^{1,0})$, we obtain \begin{align*} g(X_i,X_j)=2\operatorname{Re}h_{i\bar j}. \end{align*} Because $h$ is conjugate-linear in its second argument, \begin{align*} g(X_i,Y_j)=2\operatorname{Re}h(\partial_{z_i},i\partial_{z_j})=2\operatorname{Re}(-i h_{i\bar j})=2\operatorname{Im}h_{i\bar j}. \end{align*} Similarly, because $h$ is linear in its first argument, \begin{align*} g(Y_i,X_j)=2\operatorname{Re}h(i\partial_{z_i},\partial_{z_j})=2\operatorname{Re}(i h_{i\bar j})=-2\operatorname{Im}h_{i\bar j}. \end{align*} Finally, \begin{align*} g(Y_i,Y_j)=2\operatorname{Re}h(i\partial_{z_i},i\partial_{z_j})=2\operatorname{Re}h_{i\bar j}. \end{align*}[/step]

[guided]The goal of this step is to remove any ambiguity about the factor of $2$ and the signs. The real metric is not obtained by simply taking the real part of $h$ on the complex frame; it is obtained from the $(1,0)$-parts of real tangent vectors by the normalization \begin{align*} g(u,v)=2\operatorname{Re}h(u^{1,0},v^{1,0}). \end{align*} For the real coordinate vector fields, the $(1,0)$-parts are \begin{align*} (X_i)^{1,0}=\partial_{z_i} \end{align*} and \begin{align*} (Y_i)^{1,0}=i\partial_{z_i}. \end{align*} Thus \begin{align*} g(X_i,X_j)=2\operatorname{Re}h(\partial_{z_i},\partial_{z_j})=2\operatorname{Re}h_{i\bar j}. \end{align*} The mixed term is where the convention on complex linearity matters. Since $h$ is conjugate-linear in the second argument, scalar multiplication by $i$ in the second slot produces the factor $\overline{i}=-i$. Hence \begin{align*} h(\partial_{z_i},i\partial_{z_j})=-i h(\partial_{z_i},\partial_{z_j})=-i h_{i\bar j}. \end{align*} Taking twice the real part gives \begin{align*} g(X_i,Y_j)=2\operatorname{Re}(-i h_{i\bar j})=2\operatorname{Im}h_{i\bar j}. \end{align*} In the reversed mixed term, the scalar $i$ lies in the first argument, where $h$ is complex-linear. Therefore \begin{align*} h(i\partial_{z_i},\partial_{z_j})=i h_{i\bar j}. \end{align*} Taking twice the real part gives \begin{align*} g(Y_i,X_j)=2\operatorname{Re}(i h_{i\bar j})=-2\operatorname{Im}h_{i\bar j}. \end{align*} Finally, the two factors of $i$ cancel because the first slot contributes $i$ and the second slot contributes $\overline{i}=-i$: \begin{align*} h(i\partial_{z_i},i\partial_{z_j})=i(-i)h_{i\bar j}=h_{i\bar j}. \end{align*} Thus \begin{align*} g(Y_i,Y_j)=2\operatorname{Re}h_{i\bar j}. \end{align*}[/guided]

[step:Compute the fundamental form on the real coordinate frame] By definition, \begin{align*} \omega(u,v)=g(Ju,v) \end{align*} for real tangent vectors $u,v$. Using $JX_i=Y_i$ and $JY_i=-X_i$, the metric coefficients computed above give \begin{align*} \omega(X_i,X_j)=g(Y_i,X_j)=-2\operatorname{Im}h_{i\bar j}. \end{align*} Also, \begin{align*} \omega(X_i,Y_j)=g(Y_i,Y_j)=2\operatorname{Re}h_{i\bar j}. \end{align*} Similarly, \begin{align*} \omega(Y_i,X_j)=g(-X_i,X_j)=-2\operatorname{Re}h_{i\bar j}. \end{align*} Finally, \begin{align*} \omega(Y_i,Y_j)=g(-X_i,Y_j)=-2\operatorname{Im}h_{i\bar j}. \end{align*} [/step]

[step:Expand the proposed complex formula and compare coefficients] Define the real $2$-form $\alpha$ on $U$ by \begin{align*} \alpha=i\sum_{p,q=1}^{n}h_{p\bar q}\,dz_p\wedge d\bar z_q. \end{align*} Since $dz_p(X_i)=\delta_{pi}$, $dz_p(Y_i)=i\delta_{pi}$, $d\bar z_q(X_j)=\delta_{qj}$, and $d\bar z_q(Y_j)=-i\delta_{qj}$, direct evaluation gives \begin{align*} \alpha(X_i,X_j)=i(h_{i\bar j}-h_{j\bar i})=-2\operatorname{Im}h_{i\bar j}. \end{align*} Next, \begin{align*} \alpha(X_i,Y_j)=h_{i\bar j}+h_{j\bar i}=2\operatorname{Re}h_{i\bar j}. \end{align*} Also, \begin{align*} \alpha(Y_i,X_j)=-(h_{i\bar j}+h_{j\bar i})=-2\operatorname{Re}h_{i\bar j}. \end{align*} Finally, \begin{align*} \alpha(Y_i,Y_j)=i(h_{i\bar j}-h_{j\bar i})=-2\operatorname{Im}h_{i\bar j}. \end{align*} These four identities agree exactly with the corresponding values of $\omega$ on the frame vectors $X_i,Y_i$. [/step]

[step:Conclude equality of the forms] The real vector fields $X_1,Y_1,\dots,X_n,Y_n$ form a local frame of the real tangent bundle $TU$ over the coordinate chart. Since $\omega$ and $\alpha$ are real bilinear alternating $2$-forms and they agree on every pair of frame vectors, they agree on all real tangent vectors on $U$. Therefore \begin{align*} \omega=\alpha=i\sum_{i,j=1}^{n}h_{i\bar j}\,dz_i\wedge d\bar z_j. \end{align*} The same coefficient comparison also shows that the displayed complex expression is real as a real differential form, because all of its values on the real frame are [real numbers](/page/Real%20Numbers). This proves the claimed local formula. [/step]