TITLE: What are the exact holomorphic Lagrangians in complex 2-space?
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: I'm not sure what you mean by 'characterize'.  For example, here is a trivial characterization: Let $C$ be any Riemann surface, with meromorphic functions $a$ and $b$ with $da\not=0$.  Then let $(x,y) = (a, db/da)$.  The image of $C$ (minus a polar divisor) in $\mathbb{C}^2$ will be an algebraic curve on which $y\ dx = db$.  Thus, there are tons of these.