2
$\begingroup$

Consider a polynomial in $\mathbb{C}$ with complex coefficients, $\lambda^2+p\lambda+q$ where both $p$ and $q$ are complex numbers.

I am looking a for a condition of $p$ and $q$ such that the zeros of this polynomial would lie in the unit open disk. In other words the modulus of the zeros is strictly less than $1$.

$\endgroup$
3
  • $\begingroup$ I think that it could be $|q|\gt|p|\gt 1$. $\endgroup$
    – mike
    Commented Sep 4, 2014 at 5:30
  • $\begingroup$ @mike Thanks. Can you put some references? Or can you tell me the justification? $\endgroup$
    – Fukuzita
    Commented Sep 4, 2014 at 5:50
  • $\begingroup$ check wikipedia for "Enestrom-Kakeya Theorem". The condition is probably $q>p>1$. $\endgroup$
    – mike
    Commented Sep 4, 2014 at 6:02

2 Answers 2

1
$\begingroup$

If $p\in\mathbb{R}$ then the roots of $z^2+p z+q$ lie in the open unit disc if and only if $$ |p| < \frac{1-|q|^2}{|1-q|}.$$

Note that for $q\in\mathbb{R}$ this criterion becomes $|q|<1$ and $|p|<1+q$. If $p=|p|\omega$ for some $\omega$ on the unit circle then the criterion for real $p$ can be extended to $$|p| < \frac{1-|q|^2}{|\omega^2-q|}$$ which is valid also for complex $p$.

The denominator is bounded above by $1+|q|$. In other words $|p|+|q|<1$ will always imply that the roots will lie in the open unit disc. This latter bound also guarantees that $z^2$ and $z^2+pz+q$ have the same number of roots in the open unit disc by Rouche's theorem.

$\endgroup$
-1
$\begingroup$

I found the answer in the fantastic article written by Rahaman et.al. On the Eneström-Kakeya theorem. Tohoku Mathematical Journal, $Vol. 20 (2)$, $(1968)$, $126-136$.

Thanks to Q. I. Rahaman.

$\endgroup$
3
  • 1
    $\begingroup$ This is not useful as an answer. Could you state the actual condition? $\endgroup$
    – WimC
    Commented Sep 4, 2014 at 7:12
  • $\begingroup$ See the paper as mentioned. $\endgroup$
    – Fukuzita
    Commented Sep 4, 2014 at 7:17
  • 1
    $\begingroup$ The article can be found here. There are many bounds in that paper so its still not clear which one in particular you refer to. $\endgroup$
    – WimC
    Commented Sep 7, 2014 at 12:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .