Complex quadratic Schur criterion ($|\overline{a}b-c\overline{b}|\le |a{|}^2-|c{|}^2$)

For a complex polynomial $p(z)=a{z}^2+bz+c$, assume $|a|>|c|$ and put $D=|a{|}^2-|c{|}^2$ and $E=\overline{a}b-c\overline{b}$. Both roots of a polynomial lie in the closed unit disk if and only if $|E|\le D$. Strict inequality puts both roots in the open disk. Under the strict product assumption $|c/a|<1$, any unit-modulus root is automatically simple.

To prove the strict version, define $p^{\#}(z)=z^2\overline{p(1/\overline{z})}$ and $q=\overline{a}p-c{p}^{\#}=z(Dz+E)$. On the unit circle $|p^{\#}|=|p|$, so the Rouche theorem shows that a stable $p$ and $q$ have the same two interior zeros. Conversely $Dp=aq+c{q}^{\#}$, and the same Rouche theorem implies that two interior zeros of $q$ give two interior zeros of $p$. The nonzero root of $q$ is $-E/D$, proving the criterion. The closed version follows by continuity, replacing $b$ by $(1-\epsilon )b$ in the sufficiency direction. In the necessity direction with a boundary root, replace $p(z)$ by $p(z/r)$ for $r<1$ close to one and let $r\to 1$. The product of the two roots of a polynomial excludes two unit-modulus roots and excludes a repeated unit-modulus root.