Is there a polynomial vector field on $\mathbb{C}^2$ which possesses a bounded regular leaf? By bounded, I mean a bounded subset of $\mathbb{C}^2$.



A singular holomorphic foliation of $\mathbb{C}^2$ with a bounded leaf
Such a bounded leaf can not exist even if we assume that the vector field is analytic and the leaf is merely non-constant.


Indeed, suppose by contradiction that $i:\cal F\to \mathbb C^2$ is a leaf of an analytic vector field $v$ that is bounded subset of $\mathbb C^2$. Since $i(\cal F)$ is bounded, the vector field induces a flow on $\cal F$ that is defined for all time. So $\cal F$ has an action of $\mathbb C$ on it (without fixed points). Hence $\cal F$ is either $\mathbb C$ or $\mathbb C^\*$ or an elliptic curve. But neither of these Riemman surfaces admits a bounded holomorphic function. I.e. $i$ is constant, contradiction. 


**PS.** Maybe it's better to phrase the above in terms of the following claim.


*Statement.* Suppose by contradiction that there is a bounded leaf and $x$ is a point on it. Then there exists a holomorphic map $i:\mathbb C\to \mathbb C^2$ such that $i(\mathbb C)$ is tangent to the vector field, $i(0)=x$ and $i(\frac{\partial}{\partial z})=v$. 


This statement holds just because the flow of $v$ is defined on the leaf for all times. And it leads to a contradiction.


