Rouché’s theorem and stable root counts

Try

The polynomial z⁵+3z+1 is difficult to factor by inspection, but counting its zeros inside the radius-two disk is straightforward. On the boundary, |z⁵|=32 while |3z+1|≤7. The leading term dominates everywhere on that circle, so the full polynomial has the same five interior zeros, counted with multiplicity, as z⁵.

Rouché’s theorem answers a counting question without locating the roots. A root count is still meaningful evidence, but it is not a numerical approximation to each root.

Predict

Predict why the comparison must hold along the entire boundary, not merely at a few sampled angles. A zero could cross the boundary between samples and change the count.

Predict also whether equality is enough for the usual theorem. With f(z)=z and perturbation g(z)=−1 on the unit circle, the moduli are equal and f+g vanishes at one on the boundary. The strict inequality prevents precisely this failure.

Experiment

For p(z)=z⁴+4z+1, compare two circles.

Radius	Dominant term	Bound on other terms	Interior zero count
1	4z, modulus 4	At most 2	1
2	z⁴, modulus 16	At most 9	4

Subtracting gives three zeros in the open annulus between the circles. The strict bounds also prove that neither circle contains a zero.

Understand

Let f and g be holomorphic on a neighborhood of a simple closed contour and its interior. Suppose |g|<|f| at every contour point. Define f_t=f+tg for 0≤t≤1. The inequality |f_t|≥|f|−t|g|>0 keeps the entire deformation nonzero on the boundary.

By the argument principle, N(t)=(1/(2πi))∫f_t′/f_t dz counts the zeros of f_t inside, with multiplicity. The integrand varies continuously in t. Compactness of the contour and parameter interval, together with boundary nonvanishing, gives a uniform positive lower bound for the denominator. Therefore N(t) is continuous. Since it is integer-valued, it is constant. At t=0 it counts f; at t=1 it counts f+g. This proves Rouché’s theorem.

The theorem does not require the comparison function to be zero-free inside. Its interior zeros are exactly what we intend to count. It does require no poles in the region for this zero-only formulation; meromorphic variants need the argument principle’s zero-minus-pole accounting.

Choosing the comparison is the main practical skill. On a large circle, the highest-degree term often dominates. On a small circle, a linear or constant term may dominate. If a constant dominates, the count is zero, not one.

Apply

Find a radius on which the constant term of z³+z+5 dominates: radius one gives |z³+z|≤2<5, hence no zeros in the unit disk. Write the contour inequality before stating the conclusion.

Repair “The polynomial has five distinct zeros because the count is five.” Rouché counts multiplicity. Distinctness needs another argument, such as proving the polynomial and its derivative share no root. The theorem can count repeated zeros perfectly well without separating them.

Continue

Root counts will support local mapping arguments, while the next lessons study explicit global maps. Optional depth: show that sufficiently small coefficient perturbations preserve a polynomial’s zero count inside a fixed zero-free contour. Use the positive minimum of the original modulus on that compact contour to choose a quantitative perturbation bound.