Complex polynomial roots toy

Grab the red dots and play around! Or jump to the explanation, or try this.

Explanation

A polynomial in x of degree n has the form

From the fundamental theorem of algebra, there are exactly n roots $z_i \in \mathbb{C}$ in the complex plane.¹ The same polynomial can be written as

Let’s set $a_n=1$ for simplicity (this is called a monic polynomial).

Now, if you have the roots, finding the values of the coefficients $a_i$ is straightforward: just expand out the product in Eq. $\eqref{eq:factored}$. The functions $a_i(z_1, z_2, \ldots, z_n)$ are known as elementary symmetric polynomials (up to a sign). If you move a root, you can see how all the coefficients change.

Solving for the $z_i$’s from the $a_i$’s with some algebraic formula is possible for $n\le 4$ but generally impossible for higher degrees.² Nonetheless, we can numerically solve for roots with various numerical algorithms.³ If you move a coefficient, your computer will solve for the new locations of the roots, and you can see how they respond.

Things to try

Grab a coefficient $a_i$ and move it around in a closed loop. If it comes back to where it started, then the set of roots $\{ z_j \}$ have to return to the starting set.

But we can also talk about each individual root’s trajectory as $a_i$ is varied. If $a_i$ moves in a very small loop, so does each $z_j$.

Now try to find a larger loop for some $a_i$ so that some $z_j$’s swap places!

Hint (spoiler): a really simple choice is to move $a_0$ around the unit circle, if all the other $a_i$’s are close to the origin. Then you should see the n roots $z_j$ each shift one spot to the left/right around their unit circle. This is an n-cycle.

Try to find a 2-cycle (two roots swap places) or other more complicated types of permutations.

What we have here is a map from closed loops in a-space to permutations of the n roots.

Question: What determines the type of permutation (cycle structure or conjugacy class)? Hint: It has something to do with the zeros of the discriminant. To understand the relationship, check the box to view the discriminant’s roots. When you drag a coefficient $a_i$, you will see the roots of the discriminant (treated as univariate, holding fixed all other $a_{j\neq i}$).

Acknowledgments

This toy was somewhat inspired by John Baez’s post, which in turn was discussing this tumblr post.

This toy makes use of JSXGraph and my extended version of Polynomial.js.³

Suggestions welcome!