Evaluating a Lie Derivative of a Meromorphic Complex Form over a Holomorphic Vector Field. Given a meromorphic form $\omega : \Omega^{1, 0}$, we can get the form $\; \mathrm{d}\omega = \partial \omega + \bar \partial \omega \;$ through the use of Dolbeault operators.
I'm trying to find the lie derivative $\mathcal{L}_v \; \omega$, where $v$ is a holomorphic vector field. How can I find this for non-holomorphic forms?
For example, suppose we allow the following definitions: $$ \omega = \frac{z+4}{z-3i} \; \mathrm{d}z, \quad v = 2z-1.$$
What is the Lie derivative $\mathcal{L}_v \;\omega$?
 A: Ironically enough, I feel able enough to finally respond  with an answer.
Given complex 1-form $$\omega = \alpha(z) \mathrm \; {d}z, \qquad \alpha \in \mathcal{M}(\mathbb{C}), \qquad \frac{\partial \alpha}{\partial \bar z} = 0 \; $$ and given vector field $v = \beta(z) \; \partial_z, $ where $\beta$ is holomorphic to $\mathbb{C}$, define the Lie derivative like such:
$$\mathcal{L}_v \; \omega = i_v \circ d\omega + d \circ i_v\omega.$$
$$\mathcal{L}_v \; \omega = \; \Big(\frac{\partial \alpha}{\partial z} + \frac{\partial \alpha}{\partial \bar{z}}\Big)\beta \;\mathrm{d}\bar z + \frac{\partial(\alpha \beta)}{\partial z} \, \mathrm{d}z + \frac{\partial(\alpha \beta)}{\partial \bar z} \, \mathrm{d}\bar z.$$
Using the fact that $\frac{\partial \alpha}{\partial \bar z} = \frac{\partial \beta}{\partial \bar z} = 0$, the final result can be determined:
$$\mathcal{L}_v \; \omega = (\alpha \beta)'_z \; \mathrm{d}z + \alpha'_z \beta \; \mathrm{d}\bar{z}.$$
This is a surprisingly beautiful and simple result.
