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I would like to prove that the family of polynomials $z^{2j+2} + \alpha z^{2j+1} - \alpha z - 1$ has only one root inside the open unit circle when $|\alpha|$ is greater than 1. This seems like an ideal moment to use Rouche's Theorem, but I am having trouble finding a function to compare it to. Could anyone give me some pointers?

Antonio Vargas
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QTHalfTau
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  • For $\alpha > 0$ at least, it appears that the polynomials have all of their roots on the unit circle until $\alpha = (j+1)/j$. The polynomials only have a root inside the unit circle when $\alpha > (j+1)/j$. – Antonio Vargas Jul 13 '14 at 20:42
  • Do you have any ideas of how to prove that $|\alpha|$ must be greater than (j+1)/j ? A friend of mine believed this to be true as well, but was unable to prove anything. – QTHalfTau Jul 13 '14 at 20:57
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    Yes, I believe it should be possible using Rouché's theorem and this. I can't type up an answer right now but I'll try to return and post one within the week if someone else hasn't already. – Antonio Vargas Jul 13 '14 at 22:39
  • @AntonioVargas I might have come up with a proof, but I'm not very confident in its validity. I'd love to see a proof of yours when you have the time. – QTHalfTau Jul 14 '14 at 03:30
  • Might as well post yours. I'd be interested in seeing it! – Antonio Vargas Jul 14 '14 at 05:37
  • Okay! I'll post it after I show my friend tomorrow. I don't have a proof for the converse(that the roots are on the unit circle only if $|\alpha| \leq (j+1)/j$). However, I do have(I think) a proof that if $|\alpha| \leq (j+1)/j$ then the roots are on the unit circle. – QTHalfTau Jul 14 '14 at 13:47
  • @AntonioVargas I've found a killer error in my proof. I'm pretty much stumped at this point. – QTHalfTau Jul 15 '14 at 15:30

1 Answers1

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Let

$$ p(z) = z^{2j+2} + \alpha z^{2j+1} - \alpha z - 1 $$

and $|\alpha| > \frac{j+1}{j}$. On the circle $|z| = 1$ we calculate

$$ \left|(2j+2) z^{2j+1} - \alpha\right| \leq (2j+2) + |\alpha| < |\alpha| (2j+1) = \left|\alpha (2j+1) z^{2j}\right|, $$

so by Rouché's theorem we know that

$$ p'(z) = (2j+2) z^{2j+1} + \alpha (2j+1) z^{2j} - \alpha $$

has exactly $2j$ zeros inside the unit circle and exactly one zero outside.

When $\alpha$ is real then the coefficients of $p$ are real and

$$ -z^{2j+2} p(1/z) = p(z), \tag{$*$} $$

so by a theorem of Cohn* $p$ and $p'$ have the same number of zeros outside the unit circle (i.e. one zero). By $(*)$, if $z \neq 0$ is a zero of $p$ then so is $1/z$, so we may conclude that $p$ has exactly one zero inside the unit circle.

* See Marden, Geometry of Polynomials, section 45, Theorem (45,2).

Antonio Vargas
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  • I'm unable to prove that $p$ has all of its zeros on the unit circle when $-(j+1)/j \leq \alpha \leq -1$ and $1 \leq \alpha \leq (j+1)/j$. It can be proved for $|\alpha| < 1$ using the method in this answer but there are issues with applying Rouché's theorem in the other cases. – Antonio Vargas Jul 16 '14 at 15:58
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    Very nice. Do you think similar argument could be used to prove that $x^{n+1}+(a-1)x^n-(a+1)x+1$ has one zero in the unit circle ($a\neq 0, n \geq 2$)? I've run into this recently in https://math.stackexchange.com/questions/3596333#3598016. – Sil Mar 28 '20 at 11:15