Rouches Theorem Applied to a family of Polynomials

Question

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?

Answer 1

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.

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