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

  • 1
    $\begingroup$ Try the substitution $z\to(z+i)/(i z+1)$: this transfers zeroes on the unit circle to the real line, which is hopefully a more tractable question. $\endgroup$ Jul 14, 2014 at 23:04
  • $\begingroup$ Question: $j=\sqrt{-1}$? $\endgroup$ Jul 15, 2014 at 1:37
  • $\begingroup$ @Semiclassical Yes. Will try the circle to line transform. $\endgroup$
    – sauravrt
    Jul 15, 2014 at 2:35
  • $\begingroup$ Also, I think you can get rid of $u_0$ by doing an appropriate phase shift $z\to e^{-i \phi}$ (prior to mapping to the unit circle, I mean). So $u_0=0$ without loss of generality. $\endgroup$ Jul 15, 2014 at 2:37
  • $\begingroup$ Are you sure you want $a_n=a^*_{N-n}$? Seems we want $a_n=a^*_{N-1-n}$, since your polynomial is degree $N-1$. In particular, you didn't define $a_N$, yet your formula indicates $a_0=a^*_N$, although direct calculation seems that $a_0=a^*_{N-1}$. $\endgroup$ Jul 15, 2014 at 7:22

2 Answers 2


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

$$|a_0|\ge |a_1| \ge \cdots \le |a_{N-2}| \le |a_{N-1}|$$

enter image description here

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.

  • $\begingroup$ $|a_n|$ is of the form $1 - \beta sin(n)$ i.e. varying sinusoidally from $0 \leq n \leq N-1$. So they are not cup-shaped. As for the approach you mentioned, to ensure that $q_n(z)$ has all its zeros in or on the unit circle modulus of its coefficients have to be monotonically increasing or equal and this is satisfied when the $|a_n|$ are cup-shaped. Is my understanding correct? $\endgroup$
    – sauravrt
    Jul 15, 2014 at 14:12
  • 1
    $\begingroup$ Yes. But I got $|a_n|^2=1 + \beta^2 - 2\beta \cos((1/2)(1 + 2n - N)\pi(u_0 - u_1))$. I assume that you verified numerically that all the roots of $P(z)$ are on the unit circle. Right? $\endgroup$
    – mike
    Jul 15, 2014 at 16:32
  • $\begingroup$ Yes I did verify numerically that all roots like on unit circle. Regarding $|a_n|^2$ I was only trying to imply that its form is 1 + some sinusoidal term and that it didn't form the cup shape. $\endgroup$
    – sauravrt
    Jul 16, 2014 at 11:22
  • $\begingroup$ I see. If the coefficients are not cup-shaped, then I am not aware of any general method to prove such polynomials having onlyreal zeros. $\endgroup$
    – mike
    Jul 16, 2014 at 14:50

I assume $u_0,u_1,\beta$ are all real. For $k \in \{0,1\}$, define $\alpha_k := \sqrt{e^{j\pi u_k}}$ and define:

$$p_k(z) := \sum_{n=0}^{N-1} e^{j(n-\frac{N-1}{2})\pi u_k}z^n = \sum_{n=0}^{N-1} \alpha_k^{2n-(N-1)}z^n$$

So, $P(z) = p_0(z) - \beta p_1(z)$. Further, define:

$$r_k(z) := z^{-(N-1)}p_k(z^2) = \sum_{n=0}^{N-1} \alpha_k^{2n-(N-1)}z^{2n-(N-1)} = \frac{(\alpha_k z)^N - (\alpha_k z)^{-N}}{(\alpha_k z) - (\alpha_k z)^{-1}} = (\alpha_k z)^{1-N} \frac{(\alpha_k z)^{2N} - 1}{(\alpha_k z)^2 - 1}$$

So, $r_k(z)$ is a meromorphic function with $2N-2$ simple zeros on the unit circle, and $N-1$ poles at 0. Also, the zeros of $r_k(z)$ are square roots of the zeros of $p_k(z)$, which implies $p_k(z)$ has all its zeros on the unit circle.

The crucial feature of $r_k(z)$ is then the following property. Since $|\alpha_k| = 1$ and $z^* = z^{-1}$ on the unit circle, we have that $r_k(z) = (r_k(z))^*$ on the unit circle. That is, $r_k(z)$ is real-valued on the unit circle.

With this, consider the following. If $u_0$ and $u_1$ are not too far apart, then the zeros of $r_0(z)$ and $r_1(z)$ will alternate/interlace on the unit circle. (What "not too far apart" means here will depend on $N$. It's probably something like $|u_0-u_1| \leq \frac{2}{N}$, but don't quote me on it.) In this case, the roots of $p_0(z^2)$ and $p_1(z^2)$ then alternate as well. Finally, this implies that the roots of $p_0(z)$ and $p_1(z)$ also alternate. The rest of the argument is then a sort of Hermite-Kakeya-Obreschkoff (HKO) theorem for the unit circle.

Consider the function:

$$h(z) := \frac{p_0(z)}{p_1(z)}$$

Since $h(z^2) = \frac{r_0(z)}{r_1(z)}$ is real-valued on the unit circle, $h(z)$ therefore is as well. A standard zero counting argument (like in the proof of HKO) then implies $c_0 p_0(z) + c_1 p_1(z)$ has all its zeros on the unit circle for any real $c_0,c_1$. Therefore, $P(z) = p_0(z) - \beta p_1(z)$ has all its zeros on the unit circle.

Since HKO gives an equivalent condition for root location, this further says that if the zeros of $p_0$ and $p_1$ do not alternate (i.e., $|u_0-u_1|$ is too large), then there will be some real $\beta$ such that $P(z)$ does not have zeros only on the unit circle. However, if you are restricting $\beta$ (e.g., to $\beta > 0$), you may still have that the zeros of $P(z)$ are on the unit circle. You would have to tinker with these arguments to get something in that case.


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