Cohn's theorem

In mathematics, Cohn's theorem[1] states that a nth-degree self-inversive polynomial ${\displaystyle p(z)}$ has as many roots in the open unit disk ${\displaystyle D=\{z\in \mathbb {C}$ :|z|<1\}} as the reciprocal polynomial of its derivative.[1][2][3] Cohn's theorem is useful for studying the distribution of the roots of self-inversive and self-reciprocal polynomials in the complex plane.[4][5]

An nth-degree polynomial,

${\displaystyle p(z)=p_{0}+p_{1}z+\cdots +p_{n}z^{n}}$

is called self-inversive if there exists a fixed complex number ( ${\displaystyle \omega }$ ) of modulus 1 so that,

${\displaystyle p(z)=\omega p^{*}(z),\qquad \left(|\omega |=1\right),}$

where

${\displaystyle p^{*}(z)=z^{n}{\bar {p}}\left(1/{\bar {z}}\right)={\bar {p}}_{n}+{\bar {p}}_{n-1}z+\cdots +{\bar {p}}_{0}z^{n}}$

is the reciprocal polynomial associated with ${\displaystyle p(z)}$ and the bar means complex conjugation. Self-inversive polynomials have many interesting properties.[6] For instance, its roots are all symmetric with respect to the unit circle and a polynomial whose roots are all on the unit circle is necessarily self-inversive. The coefficients of self-inversive polynomials satisfy the relations.

${\displaystyle p_{k}=\omega {\bar {p}}_{n-k},\qquad 0\leqslant k\leqslant n.}$

In the case where ${\displaystyle \omega =1,}$ a self-inversive polynomial becomes a complex-reciprocal polynomial (also known as a self-conjugate polynomial). If its coefficients are real then it becomes a real self-reciprocal polynomial.

The formal derivative of ${\displaystyle p(z)}$ is a (n − 1)th-degree polynomial given by

${\displaystyle q(z)=p'(z)=p_{1}+2p_{2}z+\cdots +np_{n}z^{n-1}.}$

Therefore, Cohn's theorem states that both ${\displaystyle p(z)}$ and the polynomial

${\displaystyle q^{*}(z)=z^{n-1}{\bar {q}}_{n-1}\left(1/{\bar {z}}\right)=z^{n-1}{\bar {p}}'\left(1/{\bar {z}}\right)=n{\bar {p}}_{n}+(n-1){\bar {p}}_{n-1}z+\cdots +{\bar {p}}_{1}z^{n-1}}$

have the same number of roots in ${\displaystyle |z|<1.}$