**Edit 1:** For a related discussion see this [MSE post](https://math.stackexchange.com/questions/1356852/is-an-irreducible-regular-algebraic-curve-connected)


I apologize in advance, if this question is obvious:


1)What is an example of a polynomial vector field on $\mathbb{R}^{2}$ with at least two limit cycles $\gamma\_{1},\; \gamma\_{2}$ such that they lie on the same leaf of the corresponding singular holomorphic foliation of $\mathbb{C}^{2}$?


2)I have the same question by replacing "they lie on the same leaf" with "they lie on different leaves"


An indirect (or may be direct) motivation for this question is the following lemma in a celebrated paper of Petrovski and Landis "On the number of limit cycles of the equation ${dy\over dx}={P(x, y)\over Q(x, y)}$, where $P$ and $Q$ are polynomials of 2nd degree" :


"Lemma: If two distinct limit cycles of a real polynomial vector fields lie on the same complex leaf and they are homolog to each other, then the leaf is an algebraic curve"



Two limit cycles which lie on the same leaf
Possibly a very "brute force" approach could be the following. Take an non-singular algebraic curve $H(x,y)=0$ given by a polynomial $H(x,y)$ with real coefficients that has at least two ovals in the affine real plane, and generate the one-form $dH + H \omega$, where $\omega = P(x,y)dx + Q(x,y)dy$ is a real polynomial one-form. The foliation generated by the kernel of the form $dH + H \omega = 0$ gives you the orbits of the vector field
$$\dot{x}= \frac{\partial H}{\partial y}(x,y) + H(x,y) Q(x,y)$$
$$\dot{y}= - \frac{\partial H}{\partial x}(x,y) - H(x,y) P(x,y).$$ The ovals $H(x,y)=0$ are cycles of both the vector field and the form (whichever you prefer, although in the complex setting the form I think is the right object to work with). For instance, the first thing that comes to my mind is $$H(x,y) = (x^2+y^2 - 1)(x^2+2y^2-4) - 1.$$ If I am not mistaken, it is a regular curve even at infinity (it is one of these "ultra-Morse" polynomials?!). It has two nested real ovals and the rest of it is in the complex plane. I think it is a non-singular curve. If we take $\omega=y \, dx$ wouldn't that work? I haven't check whether the cycles are limit, but chances are that they are, because the form is not closed. If not, one can try a different $\omega$... 


