# Convexity of integral trajectories of rational vector field

Suppose we have a vector field determined by a rational function, of the form $$R(z) = \alpha i z + \sum_{j=1}^k \frac{c_j}{z-r_j}$$ where $$\alpha \in \mathbb{R}$$, and the other constants are in $$\mathbb{R}$$. That is, for each point $$z \in \mathbb{C}$$, $$R(z)$$ gives a complex number which we interpret as a vector. Note that with this particular choice, the integral curves, (curves following the field), look almost like large circles far away from the origin. This is due to $$\infty$$ being neither repelling nor attracting (since $$\alpha \in \mathbb{R}$$). See the attached figure for an example. Suppose now we have a closed integral curve $$\gamma$$, where all zeros and poles of $$R(z)$$ are in the interior. Let $$\gamma_0$$ be the smallest (intuitively) such integral curve.

Far away from the origin, $$\gamma$$ is more or less a circle, and thus convex. Now, as we 'shrink' $$\gamma$$, at some point (as in our figure), the integral curve cease to be convex.

Now, as $$\gamma$$ shrinks and gets more similar to $$\gamma_0$$, we go from convex to non-convex.

Is it possible that we go from non-convex back to convex again?

Phrased in a different manner: let $$\gamma$$ be a non-convex integral curve containing all zeros and poles of $$R(z)$$. Can there be a (smaller) integral curve $$\gamma^*$$ inside $$\gamma$$, which also contains all zeros and poles, but with $$\gamma^*$$ being convex?