Lemma 11.1.23B0A8DD950AD

Proof. Since $\mathrm d^{\mathrm c}f$ is a one-form, $\mathrm{d}(\mathrm d^{\mathrm c}f\wedge T)=\mathrm{dd}^{\mathrm{c}}f\wedge T-\mathrm d^{\mathrm c}f\wedge \mathrm{d}T$, and Stokes' theorem gives the first equality in

For the second, recall that the integral of $\mathrm d^{\mathrm c}\sigma $ vanishes for every form $\sigma $ of total degree $2n-1$, by Stokes' theorem for $\partial $ and $\bar\partial $. We apply this to $\sigma =f\, \mathrm{d}T$ and use $\mathrm d^{\mathrm c}\mathrm{d}=-\mathrm{d}\mathrm d^{\mathrm c}$. $\square$