# Closed orbit for vector field $f(\bar{z})$ where $f$ is holomorphic function

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?

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

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