Line bundles with complex connection

Question

Suppose that we have a complex manifold $X$, and a line bundle $L$ over $X$. It is known that the line bundles over $X$ are parametrized by their Chern class, the Chern class being the cohomology class of the curvature of a connection on $L$, which must be integral. Thus $L$ admits a connection with curvature $\omega$ iff $[\omega]$ is an integral cohomology class.
My question is this: what if we are interested instead in complex connections on $L$ (connections that are $\mathbb{C}$-linear instead of $\mathbb{R}$-linear)? Given a complex (2,0)-form $\omega$ on $X$, under what circumstances will $L$ admit a complex connection with curvature $\omega$?

Answer 1

If I understood this right, the answer to your original question is somewhat trivial. You may check that if a complex connection $\nabla$ on $L$ has curvature $\omega$, then the cohomologous $2$-form $\omega+d\alpha$ is the curvature of the connection $\nabla+\alpha$. Hence any closed $2$-form representing integral class can be the curvature form of a complex connection on a line bundle (which just means a $C^\infty$ vector bundle, not holomorphic one).

However, there is a version of your question which admits a non-trivial answer.

If you take $L$ to be a holomorphic line bundle and endow it with a Hermitian metric then there is an unique Hermitian $(1,0)$-connection called the Chern connection, its curvature form $\omega$ is a real-valued (1,1)-form (real-valued means that $\omega=\bar{\omega}$) representing an integral class.

A non-trivial result is that the converse is also true when $X$ is compact Kähler, i.e., any real-valued (1,1)-form representing an integral class is the curvature form of the Chern connection on a Hermitian holomorphic line bundle. This is a consequence of the $\partial\bar\partial$-lemma and the Lefschetz theorem on $(1,1)$-classes.

Answer 2

I interpret your line bundle $L$ to be differentiable rather than holomorphic, else the statement that line bundles are parametrized by their Chern class would be completely false.

There always exists a hermitian structure on $L$ and if you choose one, there exists a complex connexion $\nabla $ on $L$ compatible with the hermitian structure on $L$.
The curvature $\omega \in \mathcal E^{1,1}(X)$ of the connexion $\nabla $ is a $(1,1)$ form [and not a (2,0) form!] whose De Rham class computes the first Chern class of $L$ by the formula
$$ c_1(L)=[\frac {i}{2\pi}\omega ] \in H^{1,1} (X,\mathbb C)\subset H^2_{DR} (X,\mathbb C) $$

That Chern class is independent of the chosen hermitian structure on $L$ and is integral in the sense that it lives in the image of the natural group morphism (where $H^2_{sing}$ is the topologists' singular homology)

$$H^2_{sing}(X,\mathbb Z)\to H^2_{sing}(X,\mathbb C)=H^2_{DR} (X,\mathbb C)$$

NB My answer addresses the cohomology class $[ \omega] $ of the curvature. I don't know which $(1,1)$ actual forms $\omega$ you can reach by this procedure.
(You claim that you can reach any $\omega$ with integral class in the real case. I don't know why this is true).