In an exact symplectic manifold, i.e. where the symplectic form can be written $\omega = d \lambda$, it's natural to look for exact Lagrangians, i.e. $L$ on which $\lambda_L = df$. One reason is that, any disk ending on such a Lagrangian has vanishing area:

$$ \int_D \omega = \int_{\partial D} \lambda = \int_{\partial \partial D} f = 0$$

On the other hand, the search for such manifold-without-boundary Lagrangians in cotangent bundles of manifolds-without-boundary is expected to be fruitless -- Arnol'd conjectured that all such are homotopic to the zero section, and Abouzaid has proven that they're at least homotopy equivalent to the zero section.

I am interested in understanding how hard such things are to come by in the cotangent bundle of an open manifold. For cotangent bundles of one dimensional manifolds, the question is trivial -- a Lagrangian encloses area if it's a closed curve, and otherwise has no topology.

So the first case is $T^* \mathbb{C}$. What I really want to know is what the holomorphic exact Lagrangian manifolds are; this is motivated in part by the article by Xin Jin which explains that such exact holomorphic Lagrangians give rise to perverse sheaves under the Nadler-Zaslow correspondence.

All this is the motivation for the following entirely elementary question:

Characterize all algebraic curves $C \subset \mathbb{C}^2$ on which the one-form $y dx$ is exact.