A complex limit cycle not intersecting the real plane Edit:  This  is  a  real  coefficient  version  of the  current post.
Is there a polynomial vector  field $X$ with complex coefficients  on $\mathbb{C}^2$ with the property quoted bellow?

There is  a  regular leaf $L$ whose holonomy, along at least one closed curve on it,  is not trivial but $L$  does not intersect the real part $im (z)=im(w)=0,\;(z,w) \in \mathbb{C}^2$.

Note:
A leaf with non trivial holonomy is called a complex limit cycle, according to the terminology used in the  video lecture by Ilyashenko described in the following answer:
The error in Petrovski and Landis' proof of the 16th Hilbert problem
 A: A revision: Novembre  2020
I am realy indebted to Loic  Teyssier  for his $2$ very valuable  comments and  suggestions. I summarize his comments as follows:

*

*To have  a  hyperbolic complex limit cycle it is  not sufficient you check $\int_{\gamma} \alpha \neq 0$ but you should also check that this integral is different from $2k\pi i$


*If your example realy work, you can obtain a similar example with real coefficient if you replace $z^2+w^2+1=0$ with  $(z^2+w^2-4i=0)$.
Now after more than one year of his suggestions, I look at my answer again.
His first comment leads me to compute the corresponding integral again. Then I just realiz that this integral is equal to $0$!. More over I realize that not only this example is  not  appropriate for the purpose of this  question but also every possible reform of this  example is not appropriate. For example consideration of $$ \begin{cases} z'=w+z(z^2+w^2-4i)\\ w'=-z+w(z^2+w^2-4i) \end{cases}$$ does not work. For all these examples the holonomy would be tangent to the identity maps. Hence a relevant  question would be that: Are the corresponding leaf $z^2+w^2=4i$ is  a leaf with non trivial holonomy?
His second comment help me to realize that the following system has a complex limit cycle $z^2+w^2+1=0$ which obviously does not intersect the real plane $\mathbb{R}^2$. Here is the true  example as required as an answer to this post:
$$ \begin{cases} z'=w+z(z^2+w^2+1)\\ w'=-z+w(z^2+w^2+1) \end{cases}$$
Finally we include the following question in our answer:
Can a real polynomial vector field possess a hyperbolic complex limit cycle $\gamma$ which is not algebraic and does not intersect the real plane?

The previous version of my answer:
The  answer to this  question is  yes. There  is  a complex  polynomial  vector  field on $\mathbb{C}^2$ with a complex  limit  cycle  which  does  not  intersect the  real plane $im(z)=im(w)=0$.
Consider the  differential  equation $$\begin{cases}z'=w+(z^2+w^2-4i)\\ w'=-z+(z^2+w^2-4i) \end{cases}$$
The regular  leaf $L: z^2+w^2=4i$ of  this  singular  foliation does  not intersect  the  real part of  $\mathbb{C}^2$. This  leaf, which is  topologically a  cylinder,  has  a  non trivial holonomy. In  fact  we  have  more: there  is  a  closed curve on this  leaf whose  corresponding holonomy  map  is  a hyperbolic  map: namely the holonomy is  not  tangent to the identity map. Here is  the  argument:
The hyperbolicity, hence non triviality, of the  holonomy of  this  leaf is  a   consequence of  Theorem 3.2  Page 333 of  the  paper:  First  Variation of  Holomorphic  forms  and  some  applications.
Elaboration: The  foliation is defined by $$\omega= (w+(z^2+w^2-4i))dw-(-z+(z^2+w^2-4i))dz=0$$
To  apply the   theorem 3.2  in the above  paper we  find  a  $1-$ form $\alpha$ which  satisfies $d\omega=\alpha \wedge \omega$, locally around  an appropriate   closed curve $\gamma$ in $L$.
Represent  the  above  $1$- form  $\omega$ in the  form  $\omega=Pdw-Qdz$. Then for $$\alpha=(P_z+Q_w)/(P^2+Q^2)(Pdz+Qdw)$$ we  have  $d\omega=\alpha \wedge \omega$.  Note  that $P^2+Q^2$  does  not  vanish  on $L$. Now  we have to  compute $\int_{\gamma} \alpha$, along an appropriate  closed curve  $\gamma \subset L$,  and  show  that this  integral is  non zero.
To  compute  this  integral  we  parametrize the  cylinder $L$ with
$$ \phi(t)= \begin{cases} z(t)=t+i/t\\w(t)=t/i+1/t  \end{cases}$$ where  $$\phi:\mathbb{C}\setminus \{0\}\to \mathbb{C}^2$$ is the  global parametrization of $L$. We will see that the desired   appropriate  curve $\gamma$ is $\phi(S^1)$.
We  denote  by $\phi^*(\alpha) $, the pull back of $\alpha$ under embedding $\phi$. Now a  very simple  computation shows  that $\int_{S^1} \phi^* \alpha$ is  non zero  since we  obtain  a  pole of  order 1  at the  origin. In  fact  the  later  integral is  $\int_{S^1} 2(z(t)+w(t))(wdz-zdw)$. An  straightforward and  short computation shows that we  have a non degenerate  pole, namely  a  pole  of  order  1.  so the  integral  does not  vanish. So the  multiplier $e^{\int _{S^1} \alpha}  $  is  different from $1$. Then the  leaf $L$  is a hyperbolic  leaf. $\square$
