A map $f:\mathbb{C} \to \mathbb{C}$ is contractive (and Lipschitz) if for every $z_1,z_2$, we have $|f(z_1)-f(z_2)|<K|z_1-z_2|$ for some positive $K<1$. This implies the existence of a unique, attractive fixed-point.

Now, consider a set-valued map, or actually, a map $z \to Z(P(x,z)=0)$, which sends a $z \in \mathbb{C}$ to the multi-set of zeros of the polynomial $P(x,z)$.

If we are lucky, and we can factor $P(x,z)=(x-(a_1z+b_1))\dotsm (x-(a_kz+b_k))$, then I want that $z \to Z(P(x,z)=0)$ is contractive if $|a_i|<1$ for all $i$. If so, then each map $z \to a_iz+b_i$ is contractive in the sense above, and iterating each such map converges toward a unique fixed point, namely a solution of $P(z,z)=0$.

However, how can one generalize this notion of contraction for general $P(x,z)$? One property I want is that if $|z|$ is sufficiently large, then all zeros of $P(x,z)=0$ have magnitudes less than $|z|$, but I also want to capture the behavior of attractive fixed points. The attractive fixed points should be the solutions to $P(z,z)=0$.

A map $f:\mathbb{C} \to \mathbb{C}$ is contractive (and Lipschitz) if for every $z_1,z_2$, we have $|f(z_1)-f(z_2)|\le K|z_1-z_2|$ for some positive $K<1$. This implies the existence of a unique, attractive fixed-point.

Now, consider a set-valued map, or actually, a map $z \to Z(P(x,z)=0)$, which sends a $z \in \mathbb{C}$ to the multi-set of zeros of the polynomial $P(x,z)$.

If we are lucky, and we can factor $P(x,z)=(x-(a_1z+b_1))\dotsm (x-(a_kz+b_k))$, then I want that $z \to Z(P(x,z)=0)$ is contractive if $|a_i|<1$ for all $i$. If so, then each map $z \to a_iz+b_i$ is contractive in the sense above, and iterating each such map converges toward a unique fixed point, namely a solution of $P(z,z)=0$.

However, how can one generalize this notion of contraction for general $P(x,z)$? One property I want is that if $|z|$ is sufficiently large, then all zeros of $P(x,z)=0$ have magnitudes less than $|z|$, but I also want to capture the behavior of attractive fixed points. The attractive fixed points should be the solutions to $P(z,z)=0$.