Behavior of zeros of $f'$ for complex polynomials $f$ with zeros on the boundary of the unit disc.

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Suppose we have $f(z) = (z-r_1)\cdots(z-r_n)$, $|r_j| = 1$. According to the Lucas-Gauss theorem, all of the zeros of $f'$ lie in the convex hull of the $r_j$, but I discovered some behavior I don't understand while playing with the applet at http://demonstrations.wolfram.com/SendovsConjecture/.

Take, for example, the case $n = 5$, and arrange the $r_j$ in "general position" on the unit circle (so that roots of $f'$ do not coincide), taking $r_1 = 1 = e^0$. Now let $r_1$ vary continuously from $e^0$ to $e^{2\pi}$ by increasing its argument. You will find various things happening to the roots $s_1,s_2,s_3,s_4$ of $f'$; of course, when $r_1$ returns to its original position, the roots will coincide with their starting positions. However, it may be that $s_1 \mapsto s_2$, $s_2 \mapsto s_3$, $s_3 \mapsto s_4$, and $s_4 \mapsto s_1$. I have also seen $s_1 \mapsto s_1$, and $s_2 \mapsto s_3$, $s_3 \mapsto s_4$, $s_4 \mapsto s_2$.

Apparently, the act of rotating one of the $r_j$ counterclockwise one revolution is a nontrivial group action on the $s_j$. I don't understand this, or really know what to search for. What's happening here? How can I tell a priori what will happen to the $s_j$ under such an action, given the $r_j$? If this is in the literature, where can I find it?

Edit: Here's a horrible depiction of an example of order 2: An example of order 2

The blue dots are zeros of $f$; the orange dots are zeros of $f'$. The big red circle shows the motion of $r_1 = 1$, and the other four red curves show how the orange dots behave.

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Fix $4$ of the original roots, and let the $5$th move over all of $\Bbb C$. Then $P_\alpha = (X - \alpha)Q(X)$ where $Q$ is a fixed polynomial of degree $4$.

Most of the time, $P_\alpha'$ has $4$ distinct roots. Let $\{\beta_i\}$ be the roots of the discriminant $\Delta(\alpha)$ of $P_\alpha'$ (a polynomial in $\alpha$ of degree $6$, almost always). Then when $\alpha = \beta_i$, $P'\alpha$ has multiple roots. If $\alpha$ makes one small loop around one of those values, you will see two (or more, if $\beta_i$ has multiplicity) very close roots of $P'_\alpha$ being switched.

If you pick a special reference point $\alpha_0 \notin \{\beta_i\}$, and fix an ordering on the roots of $P_{\alpha_0}'$, then you can pick for each $\beta_i$ a loop from $\alpha_0$ to itself that loops only around that $\beta_i$ and describe its monodromy as a particular transposition (or a bigger cycle, in case $\beta_i$ has higher multiplicity) on the roots of $P_{\alpha_0}'$. Since those loops freely generate $\pi_1(\Bbb C - \{\beta_i\})$, you can express the original unit loop as a combination of those (up to conjugacy) and compute the corresponding permutation.


If you want a more general picture, you will need to move more roots, and ultimately, you have to describe $\pi_1(\Bbb C^5 - V(\Delta))$ (or $\pi_1((\Bbb C^5 - V(\Delta)) \cap \Bbb U^5)$ if you want to stay on the unit circle) where $\Delta$ is the discriminant of the derivative of the generic degree $5$ polynomial in term of its roots. And also, everything is prettier if you replace $\Bbb C$ with the Riemann sphere so you can discuss things as one of the root goes to infinity (this lessens the degree of $P$ by $1$).

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Let $\alpha$ be a root of $f'$. That is to say, $\alpha$ satisfies

$$ 5 \alpha^4 + 4 p_1 \alpha^3 + 3 p_2 \alpha^2 + 2 p_3 \alpha + p_4 = 0 $$

where $p_k = p_k(r_1, \dotsc, r_5)$ is a symmetric polynomial of degree $k$. When you move the $r$ around the coefficients vary polynomially and the root $\alpha$ travels along an implicit curve such that the equation above still holds. This means that $\alpha$ is again a root of the original polynomial when the parameters $r$ return to their initial position. However, this does not have to be the same root! This phenomenon is known as monodromy. Monodromy was already extensively studied by Riemann and it is the key to many beautiful results (prime examples being Riemann surfaces and his treatment of Gauss hypergeometric functions). Consider the following basic polynomial example to see more easily what is going on:

$$ \alpha^2 - r = 0. $$

Now start at $\alpha = r = 1$ and rotate $r$ once along the unit circle. You will end up with $\alpha = -1$, which is the other root of the quadratic equation $\alpha^2-1=0$.