I'm trying to prove a self-inversive polynomial $P(z) = \sum\limits_{n=0}^{N-1}a_nz^n$ has all its roots on the unit circle. The coefficients are such that
$ a_n = e^{j(n-\frac{N-1}{2})\pi u_0} – \beta e^{j(n-\frac{N-1}{2})\pi u_1}$ and $0 \leq n \leq N – 1$
These coefficients satisfy $a_n = a^*_{N-1-n}$ i.e. $P(z)$ is self-inversive.
The necessary and sufficient condition for a self-inversive polynomial to have all roots on the unit circle is that $P'(z)$ has all its roots in $|z| \leq 1$.
I considered Eneström–Kakeya theorem to show $P'(z)$ has all its roots inside unit circle, but the theorem extended for complex polynomial doesn't seem to be valid for the above polynomial.
I'm unable to make headway in trying to prove $P(z)$ has all roots on unit circle although numerical experiments show the roots are on unit circle and infact roots of $P'(z)$ are inside the unit disk.
Please provide me with any suggestions on how to approach the proof.
Thanks
Best Answer
This is not a answer. I just want to use the space and easy-to-edit feature in the "Answer" section to type the equations.
You did not specify the relation of the modulus of the coefficients $a_n$. If they happen to be cup-shaped, then you may use a theorem by Chen (J. of Math. Anal. and Appl. vol 190, 714-724 (1995)).
By cup-shaped, I mean
$$|a_0|\ge |a_1| \ge \cdots \le |a_{N-2}| \le |a_{N-1}|$$
In your case, we have (assume $N=2n+2$) $$P(z) = \sum_{k=0}^{2n+1}a_kz^k=z^n q(z)+q^*(z)$$
where $$q_n(z)=\sum_{k=0}^na_{n+k+1}z^k$$ $$q_n^*(z)=\sum_{k=0}^n a_kz^k$$
If $N=2n+1$, then we define $Q(z)=(1+z)P(z)$ and treat $Q(z)$ in a similar fashion.