Degree of a Normalized Polynomial on the Boundary

From a Polynomial to a Circle Map

A complex polynomial can be large and complicated on a circle, but its highest-degree term controls its behavior on circles of sufficiently large radius. The useful topological step is to normalize the polynomial’s values to have modulus one. Once the polynomial is known not to vanish on the circle, this gives a map from one circle to another. We will show that its degree is the degree of the polynomial, while its leading coefficient contributes only a constant phase.

We give \(\mathbb{C}\) its usual topology and each circle its subspace topology. Write \(S^1=\{\zeta\in\mathbb{C}:|\zeta|=1\}\). For \(R>0\), the circle of radius \(R\) is \(C_R=\{z\in\mathbb{C}:|z|=R\}\), oriented by the parametrization \(t\mapsto R e^{it}\) as \(t\) increases. Let \(p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0\), where \(n\geq1\) and \(a_n\neq0\). The normalization is defined at \(z\) exactly when \(p(z)\neq0\).

The condition that \(p\) have no zeros on the circle will follow from a simple estimate. For \(|z|=R\), the leading term has modulus \(|a_n|R^n\), while the sum of the other terms has modulus at most \(\sum_{k=0}^{n-1}|a_k|R^k\). Thus a strict inequality between these quantities prevents cancellation.

Proof. For \(|z|=R\), the triangle inequality gives \(|p(z)|\geq |a_nz^n|-\left|\sum_{k=0}^{n-1}a_kz^k\right|\). Applying the triangle inequality again to the lower-degree sum gives \(\left|\sum_{k=0}^{n-1}a_kz^k\right|\leq\sum_{k=0}^{n-1}|a_k|R^k\). Consequently \(|p(z)|\geq |a_n|R^n-\sum_{k=0}^{n-1}|a_k|R^k>0\), proving the first claim. For the second, divide the proposed strict inequality by the positive number \(|a_n|R^n\). Its left side becomes \(\sum_{k=0}^{n-1}(|a_k|/|a_n|)R^{k-n}\). Each term tends to zero as \(R\) tends to infinity, since \(k-n<0\), and there are only finitely many terms. The sum therefore tends to zero, so it is less than \(1\) for all sufficiently large \(R\). This is the required condition. \(\square\)

The Degree of the Normalized Boundary Map

Proof. Fix such an \(R\). For \(s\in[0,1]\), define \(P_s(z)=a_nz^n+s\sum_{k=0}^{n-1}a_kz^k\). If \(|z|=R\), then \(|P_s(z)|\geq |a_n|R^n-s\sum_{k=0}^{n-1}|a_k|R^k\geq |a_n|R^n-\sum_{k=0}^{n-1}|a_k|R^k>0\). So the formula \(H_s(\zeta)=P_s(R\zeta)/|P_s(R\zeta)|\) defines a continuous homotopy of maps \(S^1\to S^1\). At \(s=1\) it is \(f_R\); at \(s=0\) it is the normalized leading monomial. Homotopy invariance of degree, established in the earlier material on degree and winding number, gives \(\deg(f_R)=\deg(H_0)\). Write \(a_n/|a_n|=e^{i\theta}\) for some real \(\theta\). Then \(H_0(e^{it})=(a_nR^ne^{int})/(|a_n|R^n)=e^{i(nt+\theta)}\). Thus \(F(t)=nt+\theta\) is a lift, and \(F(t+2\pi)-F(t)=n(t+2\pi)+\theta-(nt+\theta)=2\pi n\). By the definition of degree, \(\deg(H_0)=n\), hence \(\deg(f_R)=n\). The constant phase \(\theta\) cancels from the lift increment, so the leading coefficient does not alter the degree. \(\square\)

This argument separates two roles. The estimate proves that normalization is allowed throughout the homotopy; homotopy invariance then transfers the degree from the polynomial to its leading term. The strict inequality matters: merely knowing that the leading term is at least as large as the sum of the remaining terms would not rule out a zero through exact cancellation.

What the Radius Condition Does—and Does Not—Say

The theorem applies to every radius satisfying the stated dominance condition, and that condition holds for all sufficiently large radii. It does not assert that every circle gives degree \(n\). For example, let \(p(z)=z-1\). On the circle of radius \(1/2\), the leading-term modulus is \(1/2\) and the lower-term bound is \(1\), so the dominance condition fails. In fact, the normalized map on that circle has degree \(0\): its values come from \(-1+(1/2)e^{it}\), and the homotopy \(-1+s(1/2)e^{it}\) never vanishes because its modulus is at least \(1-s/2\geq1/2>0\). At \(s=0\) its normalized map is constant, of degree \(0\). On the circle of radius \(2\), the leading-term modulus is \(2\), the lower-term bound is \(1\), and the theorem gives degree \(1\).

Common pitfall. The leading coefficient’s phase is not an additional contribution to degree. For the normalized monomial \(-iz^2\) on the unit circle, a mistaken calculation might assign degree \(-1/4\) to the phase \(-\pi/2\), or try to add that phase directly to the degree. But a lift is \(F(t)=2t-\pi/2\), and its increment is \(F(t+2\pi)-F(t)=(2t+4\pi-\pi/2)-(2t-\pi/2)=4\pi\). The degree is \(4\pi/(2\pi)=2\). A constant phase shifts the lift but cancels when its increment is computed.

The central check before normalizing is always that the polynomial is nonzero at every point of the boundary circle. Once a strict dominance estimate verifies this, the homotopy reduces the calculation to the leading monomial, where the lift displays the degree directly.