Friday, January 20, 2012

Invariant polynomials and geometric transformations

A couple days ago, I stumbled over a really interesting problem looking for something to post for #ProblemOfToday. It was a problem that appeared in the 1989 Putman exam and reads as follows

"Prove that if 
$11z^{10}+10 i z^9 + 10 i z -11 =0$
then $|z|=1$."

It is a really nice problem in itself, but after thinking a bit on it, I thought what was so special about this specific coefficients to have this nice property. First, the polynomial can be analyzed in an even nicer way. By doing the transformation $z\mapsto i z$ we can see that the polynomial gets mapped (up to a negative sign) to:


In this form one can see better how is the dependence of the polynomial on one of the coefficients (say 10), and can quickly ask a generalization of this particular problem:

$(n+1)z^n+nz^{n-1}+n z+ (n+1)=0$
then $|z|=1$ for $n\in\mathbb{Z}/\{0\}$.

Doing a couple of special cases, one can get convinced that actually the previous statement holds true. 



In the general case maybe the answer relies in a geometric argument. The polynomials $p_n(z)=(n+1)z^n+nz^{n-1}+n z+ (n+1)$ have the peculiarity that their roots are invariant under the  inversion of the complex plane, i.e. by doing the transformation $z \mapsto \frac{1}{z}$, we have that
that is, roots are mapped into roots by the inversion. 

Therefore, if there is a root bigger than 1, there should be a root smaller than 1 and vice-versa. It is not difficult to see that here cannot be a root bigger than 1, as $(n+1)z^n+nz^{n-1}$ and $n z+ (n+1)$ would have to be equal in modulus, but their orders of magnitude are different. 

Another way of proving this is by analyzing $p_n\left( e^{i \theta}\right)$. After a little simplification we have that $p_n\left( e^{i \theta}\right)=2e^{\frac{i \theta n}{2}}\left( n\cos\left( \frac{(n-2)\theta}{2}\right)+(n+1)\cos\left( \frac{n\theta}{2}\right)\right)$, which can be found to have exactly $n$ real roots for $\theta$. Hence all roots of $p_n$ lie on the unit circle.

The interesting fact is that for $n\to\pm\infty$, the roots of $p_n(z)$ become dense on $S^1$
For example, the graph of the absolute value of $p_21(z)$ is given by

where the $S^1$ can be seen. Likewise, for negative values of $n$ we can recover $S^1$

It looks like the $p_n(z)$ can be thought as some orthogonal polynomials whose support is $S^1$, and for the same reason, it is also natural to think that they can be eigenfunctions (up to a renormalization) of some operator (hopefully differential!).

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