Suppose p(z) is a monic, n degree polynomial with no roots.
The term zn dominates, for large enough z, so there exists some radius r such that every z beyond r has p(z) nonzero. For instance, r can be set to the sum of the norms of the coefficients. Since p is monic, r is at least 1.
Let f = p(z)/|p(z)| on the circle of radius r. The denominator is nonzero, so this is well defined. And it's continuous from S1 to S1.
Apply a linear homotopy that shrinks all the coefficients, except for the lead coefficient, down to 0. We are left with the function zn, which wraps the circle around itself n times. Therefore our original function f has degree n.
We're assuming p has no roots, so the same formula, p(z)/|p(z)|, is well defined throughout the disk of radius r, and it's continuous. In the previous section, we showed that a function of degree n cannot be extended continuously to a disk, thus we have a contradiction, and p(z) has a root after all.
when a root w is found, divide p(z) by z-w to give a lower degree polynomial. Repeat this process until you have n roots, some of which may be duplicates.
Split p(z) into n linear factors z-wi. This splits f into the product of z-wi over |z-wi|. This is the angle from wi out to z, as z traces the unit circle. For instance, if wi = 0, z/|z| is simply z. The angle sweeps through 360 degrees, and the map has degree 1. In fact, the angle of z-wi makes a complete revolution for any root wi inside the unit circle, giving a map of degree 1.
If wi lies outside the unit circle, the angle of z-wi moves back and forth, as z runs around the circle, but it always stays within a range of 180 degrees. This map has degree 0.
The factor associated with the root wi has degree 1 or 0, when wi is inside or outside the unit circle respectively.
Demoivre's theorem says that angles are added as complex functions are multiplied. Therefore f = p(z)/|p(z)| is the sum of the angles associated with the linear terms z-wi. If we go back to the definition of degree, the degree is the difference between the end points of the lift into R1. When angles are added downstairs, their lifts are added upstairs, and the degrees are added as well. Therefore the degree of f is the number of roots inside the unit circle.
Let m be the minimum of |g| on the unit circle. since |g| attains its minimum somewhere on the circle, m is positive.
Build a power series for g. Since g is analytic throughout the unit disk, the power series converges uniformly to g on the disk. Let p(z) be the tailor approximation that is within m/2 of f everywhere on the unit disk.
A homotopy starts with g, and shrinks all the remaining terms in the power series to 0, leaving p(z). The same homotopy carries f to p(z)/|p(z)|, which is well defined, since p(z) is never 0 on the unit circle. Thus f has the same degree as p(z)/|p(z)|, which is the number of roots inside the unit circle. The degree of f is nonnegative.