Roots of $g(z) - g(e^{i \theta})$ in terms of $\theta$ Given a complex polynomial
$$g(z) = \sum_{n=0}^N c_n z^n $$
I'd like to express the roots of $f(z) = g(z) - g(e^{i \theta})$ as some function of $\theta$.
I've been experimenting with graphing the path of the roots as $\theta$ changes and the result looks like a smooth curve.  Moreover, the roots actually trace paths which sometimes meet up with the paths from other roots to form larger paths.  So it looks like the roots are evenly spaced points on one or more complex curves.
For instance, for $g(z) = .16z^3 + \frac{1}{2}z^2 + z$, the trace looks like (the black circle is the unit circle, for scale):

So I'm guessing there's some nice form for the paths the roots trace and their positions on these paths, but I'm not sure how to begin approaching this problem.
 A: Here's a contribution to the overall problem, too big to fit into a comment. Maybe this can inspire a real proof by someone else.
I think it'll be pretty easy to show that the paths are continuous, so long as you can define the path in a way that accurately "traces" one of the roots around. To this end, let $f:\mathbb{C}\to\mathbb{C}$ be a polynomial. Now fix $z_0\in\mathbb{C}$ and let $c=f(z_0)$. I claim that for every real $\epsilon>0$ there exists $\delta>0$ such that for every $c_1\in\mathbb{C}$, if $|c_1-c|<\delta$, then there exists $z_1\in\mathbb{C}$ with $|z_1-z_0|<\epsilon$ such that $f(z_1)=c_1$. In other words, if we have a solution for $f(z)=c$, then if $c_1$ is not too far away from $c$ then we can find a solution $f(z_1)=c_1$, where $z_1$ is as close as we like to $z$. If $z$ is a single root of $f-c$, then this proof is as easy as considering the inverse, which must be continuous. If $z$ is a multiple root, it's more difficult but I think still doable.
Assuming that my continuity argument is correct, we can say that there are well-defined paths, i.e. there are no jumps. We need to take care with multiple roots, making sure they don't stray onto each other's paths, but I think it's doable.
But each root has a continuous path, and there has to be some sort of parametrization that, with a little care, we can show is periodic. Defining $g:\mathbb{R}^2\to\mathbb{C}$ as $g(x,y)=f(x+yi)$, we can then use the continuity and (real) differentiability of the gradient of $g$ combined with information on how $e^{i\theta}$ changes to show that there's a smooth path.
Do I have anything here, or is it all a load of nonsense? Let me know! :)
