A polynomial having same modulus that of an associated polynomial on the imaginary axis

Question

Let $P(z)$ be a complex polynomial of degree $n.$ I am working on the class of polynomials assiociated to $P(z)$ such that their moduli are identical with that of $P(z)$ on the imaginary axis.

For example if $Q(z)$ is a polynomial obtained by the replacement of coefficients of $P(z)$ by their complex conjugates and $z$ is replaced by $-z,$ then $|P(iy)|=|Q(iy)| $ where $y$ is any real.

The above is one such associated polynomial. May I request you to share your thoughts on this class of polynomials with respect to $P(z).$ Do there exist any other such polynomials which behave similarly on the imaginary axis?

Answer 1

The set of all polynomials associated with a given one is described as follows: Let the given polynomial be $$P(z)=c(z-z_1)\ldots(z-z_n).$$ Then any associated polynomial is of the form $$Q(z)=\lambda c(z-\sigma_1(z_1))\ldots(z-\sigma_n(z_n)),$$ where each $\sigma_j(z)=z$ or $-\overline{z}$, and $|\lambda|=1$. So, besides the continuous parameter $\lambda$ you have at most $2^n$ possible sets of zeros in a group of associated polynomials ($\sigma$ does not change imaginary roots).

The proof is simple. Let $P,Q$ be associated, and assume for simplicity that they are monic (the coefficients at top degree are equal to $1$). Then the polynomials $P(z)\overline{P(-\overline{z})}$ and $Q(z)\overline{Q(-\overline{z})}$ have the same zeros and are monic, and coincide on the imaginary axis (because they are both non-negative and have equal absolute values on the imaginary axis). Therefore they are equal. So each zero of $Q$ is either a zero of $P$ or symmetric to a zero of $Q$ with respect to the imaginary axis.