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Let $(M,J)$ be a complex manifold ,prove the following result :complex vector field $Z$ is of $(0,1)$ type if and only if $Z(f) = 0$ for every locally holomorphic function $f$.

My attempt:

first by the definition or the dual version,$Z$ is of type $(0,1)$ if and only if for all $(1,0)$ form $\omega$ ,we have $\omega(Z) = 0$ ,in particular $Z(f) = df(Z)$ ,observe that $d = \partial +\bar\partial$ since $f$ is holomorphic $df = \partial f$ is of type $(1,0)$ hence $Z(f) = 0$.

Conversely,since any $(1,0)$ form locally can be written as linear combination of $dz_i$ while each $z_i$ are locally holomorphic function ,hence $\omega(Z) = c_idz_i(Z) = c_iZ(z_i) = 0$,which implies $Z$ is of type $(0,1)$.Is my proof correct?Is there some better proof?

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