Schur stable polynomial

Remarks

Example: $P(z) = z^2 - z + \tfrac{1}{2}$ has roots $\tfrac{1}{2} \pm \tfrac{1}{2} i$. Each has modulus $\sqrt{1/4 + 1/4} = 1/\sqrt{2} < 1$, so $P$ is Schur stable.

Non-example: $P(z) = z^2 - 1$ has roots $1$ and $-1$. They have modulus exactly $1$, so the strict inequality fails and $P$ is not Schur stable.

Some authors call such polynomials simply stable in a discrete-time context, or say that $P$ has all its zeros in the open unit disc. The polynomial $P$ is Schur stable if and only if its reversed polynomial $z^n P(1/z)$ has all its roots outside the closed unit disc, provided $a_0 \ne 0$.