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Edit : According to the comments of Michael Renardy and Christian Remling I revise the question as follows:

Is there a vector field $X$ on an open set $U\subseteq \mathbb{R}^2$ such that $X $ has a closed orbit and is in the form $X=f(\bar{z})$ where $f$ is a holomorphic function on $\overline{U}=\{\bar{z}\mid z\in U\}$?

Added after the answer by Prof. Duchon: Is there an example of such vector field with an Isochronous band of closed orbits?

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    $\begingroup$ Wouldn't such a vector field be divergence free? $\endgroup$
    – Michael Renardy
    Commented Sep 9, 2017 at 17:01
  • $\begingroup$ @MichaelRenardy Yes of course. Thank you for your very interesting point. My apology for my elementary question. $\endgroup$
    – Ali Taghavi
    Commented Sep 9, 2017 at 17:10
  • $\begingroup$ @MichaelRenardy Is there an example of such vector field with a band of closed orbits on for a (non simply connected ) open set $U$ ? $\endgroup$
    – Ali Taghavi
    Commented Sep 9, 2017 at 17:41
  • $\begingroup$ @ChristianRemling Yes, Thank you. I understand you are saying that two vector fields $f(\bar{z})$ and $\overline{f(z)}$ are smoothly equivalents. $\endgroup$
    – Ali Taghavi
    Commented Sep 9, 2017 at 17:55

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Yes, $U=\{z:a<|z|<b\}$ ($a>0$) and $f(z)=i/\bar z$. Orbits are circles $\{|z|=c\}$.

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  • $\begingroup$ Dear Prof. Duchon. Thank you so much for this answer. Is there an example of such vector field with a band of closed orbits with constant period? $\endgroup$
    – Ali Taghavi
    Commented Sep 9, 2017 at 19:06
  • $\begingroup$ May I have your email address? My I ask you that you kindly send me a message to my email address "[email protected]"?Thank you. $\endgroup$
    – Ali Taghavi
    Commented Sep 9, 2017 at 19:12

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