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5 typo

The 2-variable polynomial equation $f(z,t) = 0$ with $z = \mathbb{C}, \,t \in \mathbb{S^1}$ has $n = \mathrm{deg}_z f$ solutions each fixed $t$. I wanted to follow the roots as they travel with time paramter $t$ and count the number of distinct orbits. Is there a procedure for this?

Example: $f(z,t) = z^3 t^2 + z^2 t + z t^3 + z$ as our "time" paramter parameter moves in a circle, the roots follow 2 distinct orbits. Here, $t \in \{ 0.9 e^{i \theta}: \theta \in [0, 2 \pi] \} = 0.9 S^1 \subset \mathbb{C}$

This defines a closed path in the configuration space of three points in the complex plane, i.e. a braid. My plot is a projection of this braid, forgetting the crossings. Looks like an circle + trefoil (not linked).

EDIT: One point of the fundamental theorem of algebra is smooth deformations in the space of polynomials will not change the number of roots (if we projectivize). Proofs always involve deformations from arbitrary polynomials to $p(z) = z^n$ for some natural number degree $n = \deg p$.

If we add a periodic "time" variable, I expected each root to follow a disjoint closed path, but instead paths join together sometimes. So I started to ask about the number of components.

Geometrically, I would like to know if you can deform $\{ f(z,t)=0 \} \in \mathbb{C} \times S^1$ without changing the number of components. From the comments, it sounds like the number of components is constant unless the discriminant vanishes and whether it takes values in the unit circle. As long as these two events do not happen, is the topology of this set "constant"?

As an example: $f(z,t) = 1 + 3 z t + z^2 t + z^3 (2 + t)$ can be deform to $f_0(z,t) = z^3 + t z$ by "turning off" various coefficients. The number of components will be 2.

COMMENT: The closest thing I could find is the work of Vivek Shende and Alexei Oblomkov, where they study the intersection $\{ f(z,w) = 0 \} \cap \{ |z|^2 + |w|^2 = \epsilon\} \subset \mathbb{C}^2$ with $\epsilon << 1$ and $f(z,w)$ singular at $(z,w) = (0,0)$.

Topologically, it is a link in a 3-sphere. Generically it's trivial though it's possible to get torus knots for $f(z,w) = z^p + w^q$.

Comments in there point in various directions, e.g. Three-Dimensional Link Theory and Invariants of Plane Curve Singularities but it's on a different "setup".

4 added another example

The 2-variable polynomial equation $f(z,t) = 0$ with $z = \mathbb{C}, \,t \in \mathbb{S^1}$ has $n = \mathrm{deg}_z f$ solutions each fixed $t$. I wanted to follow the roots as they travel with time paramter $t$ and count the number of distinct orbits. Is there a procedure for this?

Example: $f(z,t) = z^3 t^2 + z^2 t + z t^3 + z$ as our "time" paramter moves in a circle, the roots follow 2 distinct orbits. Here, $t \in \{ 0.9 e^{i \theta}: \theta \in [0, 2 \pi] \} = 0.9 S^1 \subset \mathbb{C}$

This defines a closed path in the configuration space of three points in the complex plane, i.e. a braid. My plot is a projection of this braid, forgetting the crossings. Looks like an circle + trefoil (not linked).

EDIT: One point of the fundamental theorem of algebra is smooth deformations in the space of polynomials will not change the number of roots (if we projectivize). Proofs always involve deformations from arbitrary polynomials to $p(z) = z^n$ for some natural number degree $n = \deg p$.

If we add a periodic "time" variable, I expected each root to follow a disjoint closed path, but instead paths join together sometimes. So I started to ask about the number of components.

Geometrically, I would like to know if you can deform $\{ f(z,t)=0 \} \in \mathbb{C} \times S^1$ without changing the number of components. From the comments, it sounds like the number of components is constant unless the discriminant vanishes and whether it takes values in the unit circle. As long as these two events do not happen, is the topology of this set "constant"?

A more refined question would

As an example: $f(z,t) = 1 + 3 z t + z^2 t + z^3 (2 + t)$ can be deform to ask about the braid$f_0(z,t) = z^3 + t z$ by "turning off" various coefficients. The number of components will be 2.

COMMENT: The closest thing I could find is the work of Vivek Shende and Alexei Oblomkov, where they study the intersection $\{ f(z,w) = 0 \} \cap \{ |z|^2 + |w|^2 = \epsilon\} \subset \mathbb{C}^2$ with $\epsilon << 1$ and $f(z,w)$ singular at $(z,w) = (0,0)$.

Topologically, it is a link in a 3-sphere. Generically it's trivial though it's possible to get torus knots for $f(z,w) = z^p + w^q$.

Comments in there point in various directions, e.g. Three-Dimensional Link Theory and Invariants of Plane Curve Singularities but it's on a different "setup".

EDIT: One point of the fundamental theorem of algebra is smooth deformations in the space of polynomials will not change the number of roots (if we projectivize). Proofs always involve deformations from arbitrary polynomials to $p(z) = z^n$ for some natural number degree $n = \deg p$.
Geometrically, I would like to know if you can deform $\{ f(z,t)=0 \} \in \mathbb{C} \times S^1$ without changing the number of components. From the comments, it sounds like the number of components is constant unless the discriminant vanishes and whether it takes values in the unit circle. As long as these two events do not happen, is the topology of this set "constant"?
COMMENT: The closest thing I could find is the work of Vivek Shende and Alexei Oblomkov, where they study the intersection $\{ f(z,w) = 0 \} \cap \{ |z|^2 + |w|^2 = \epsilon\} \subset \mathbb{C}^2$ with $\epsilon << 1$ and $f(z,w)$ singular at $(z,w) = (0,0)$.