Inspired by this question and the counter example provided in its answer we ask:

Is there a polynomial vector field on $\mathbb{R}^2$ such that after complexification of the equation, the corresponding singular holomorphic foliation of $\mathbb{C}^2$ possess a regular complex leaf $L$ whose holonomy is nontrivial and $L$ does not intersect the real part of $\mathbb{C}^2$? That is $L$ does not intersect $\{(z,w)\in \mathbb{C}^2 \mid im(z)=im (w)=0\}$.

Note: The counter example in the above linked post shows that this situation can occur if the coefficient of polynomial vector field are complex. But what about if the coefficients are real?

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    $\begingroup$ This case is the same as your example in the complex version of the question. Just consider $$z′=w+(z^2+w^2+4) \\ w′=−z+(z^2+w^2+4)$$. $\endgroup$ Apr 25, 2019 at 14:10
  • $\begingroup$ @LoïcTeyssier Yes. Many thanks for this comment. $\endgroup$ Apr 26, 2019 at 7:25
  • $\begingroup$ @LoïcTeyssier Now I think that the example i provided in the linked question is somewhat a fake example, in the sense that the leaf is an algebraic leaf. now I am thinking to find an algebraic vector field with a non algebraic complex limit cycle not intersecting the real plane(both real or complex coefficents). As we know a generic algebraic vector field does not have an algebraic leaf.So in this new formulation is the question still an obvious question? $\endgroup$ Apr 26, 2019 at 8:40
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    $\begingroup$ I'm not convinced that the algebraic nature of the leaf is relevant here. But I already was wrong on this question ;) $\endgroup$ Apr 28, 2019 at 19:14
  • $\begingroup$ @LoïcTeyssier No but your comment was interesting. May be it is an indirect motivation to consider the following example: $\begin{cases}x'=y-x^2\\y'=-x \end{cases} $ Now what about the curve $e^{-2y}(y-x^2+1/2)=i$? What can be said about its holonomy?Is it nontrivial? I appreciate your comments to this question. $\endgroup$ May 2, 2019 at 10:34

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