Why the Leading Monomial Controls a Large Circle
A polynomial can have complicated lower-degree terms, but on a sufficiently large circle its highest-degree term dominates their combined contribution. That comparison gives more than a pointwise estimate: it allows us to move continuously from the polynomial to its leading monomial while keeping every value nonzero. The nonvanishing condition is essential when the homotopy is viewed as a homotopy of maps into the punctured plane.
Write \(\mathbb{C}\) with its usual topology and modulus, and give every circle below the subspace topology inherited from \(\mathbb{C}\). For \(R>0\), write \(C_R=\{z\in\mathbb{C}:|z|=R\}\). A homotopy from \(f:C_R\to\mathbb{C}\setminus\{0\}\) to \(g:C_R\to\mathbb{C}\setminus\{0\}\) is a continuous map \(H:[0,1]\times C_R\to\mathbb{C}\setminus\{0\}\) such that \(H(0,z)=f(z)\) and \(H(1,z)=g(z)\) for every \(z\in C_R\). The interval and product here have their usual topologies.
The Straight-Line Homotopy
For a radius satisfying strict dominance, define $$ H(t,z)=(1-t)p(z)+t\,a_nz^n,\qquad t\in[0,1],\ z\in C_R. $$ At \(t=0\) this is \(p(z)\), and at \(t=1\) it is \(a_nz^n\). The key point is to prove that no intermediate value is zero.
Proof. For \(z\in C_R\), separate the leading term by writing $$ p(z)=a_nz^n+\sum_{k=0}^{n-1}a_kz^k. $$ Consequently the proposed interpolation can also be written $$ H(t,z)=a_nz^n+(1-t)\sum_{k=0}^{n-1}a_kz^k. $$ The reverse triangle inequality and \(0\leq 1-t\leq1\) give $$ |H(t,z)|\geq |a_nz^n|-(1-t)\left|\sum_{k=0}^{n-1}a_kz^k\right| \geq |a_n|R^n-\sum_{k=0}^{n-1}|a_k|R^k>0. $$ The second inequality uses the triangle inequality and \(|z^k|=R^k\) on \(C_R\). Thus \(H(t,z)\) never equals zero. It is continuous because it is given by sums and products of continuous functions of \(t\) and \(z\); since its values are nonzero, it is a continuous map into the subspace \(\mathbb{C}\setminus\{0\}\). Finally, substituting \(t=0\) gives \(H(0,z)=p(z)\), and substituting \(t=1\) gives \(H(1,z)=a_nz^n\). This is the required homotopy. \(\square\)
The proof provides a uniform lower bound: on the entire circle, at every time \(t\), the modulus of \(H(t,z)\) is at least the positive difference between the leading-term bound and the sum of the lower-term bounds. Thus it is not enough that the two endpoint maps avoid zero separately; the estimate controls every point of the interpolation.
Why a Sufficiently Large Radius Exists
Proof. First suppose the degree \(n\) is positive. For each index \(k<n\) with \(a_k\ne0\), the inequality $$ |a_k|R^k<\frac{|a_n|R^n}{2n} $$ is equivalent, after division by the positive number \(|a_k|R^k\), to $$ R^{n-k}>\frac{2n|a_k|}{|a_n|}. $$ There are only finitely many nonzero lower coefficients. Choose \(R>0\) larger than each of the finitely many positive thresholds \(\bigl(2n|a_k|/|a_n|\bigr)^{1/(n-k)}\). If there are no nonzero lower coefficients, any positive \(R\) works. For the chosen \(R\), each nonzero lower term is strictly less than \(|a_n|R^n/(2n)\), and zero coefficients contribute zero. There are at most \(n\) lower terms, so $$ \sum_{k=0}^{n-1}|a_k|R^k< n\frac{|a_n|R^n}{2n}=\frac{|a_n|R^n}{2}<|a_n|R^n. $$ This proves strict dominance. If \(n=0\), the sum is empty and equals zero, while \(|a_0|R^0=|a_0|>0\); hence the inequality holds for every \(R>0\). \(\square\)
For a constant polynomial, the homotopy has no actual movement: \(p(z)=a_0\) is already its leading monomial. The proposition includes this case so that the statement does not silently exclude degree zero. The zero polynomial is different: it has no leading coefficient and no leading monomial, so the claim is not made for it.
Worked Example: A Real-Coefficient Cubic
Take \(p(z)=2z^3+z^2-3z+1\) and \(R=2\). The leading term has modulus $$ |2z^3|=2|z|^3=2\cdot 2^3=16. $$ The sum of the lower-term bounds is $$ |z^2|+|-3z|+|1|=2^2+3\cdot2+1=4+6+1=11. $$ Since \(16>11\), strict dominance holds. The interpolation is $$ H(t,z)=2z^3+(1-t)(z^2-3z+1). $$ For every \(z\in C_2\) and \(t\in[0,1]\), the reverse triangle inequality gives $$ |H(t,z)|\geq16-(1-t)(4+6+1) =16-11(1-t)\geq16-11=5>0. $$ At \(t=0\), substitution gives \(2z^3+z^2-3z+1\); at \(t=1\), it gives \(2z^3\). Check: the uniform lower bound is \(5\), so even the smallest allowed bound is positive; the homotopy never reaches zero.
Worked Example: Complex Coefficients
Let \(p(z)=(1+i)z^2+(1-i)z+1\) and take \(R=2\). The leading-term modulus is $$ |(1+i)z^2|=|1+i|\cdot 2^2=\sqrt{2}\cdot4=4\sqrt{2}. $$ The lower-term bound is $$ |(1-i)z|+|1|=\sqrt{2}\cdot2+1=2\sqrt{2}+1. $$ Their difference is \(4\sqrt{2}-(2\sqrt{2}+1)=2\sqrt{2}-1>0\), since \(2\sqrt{2}>1\) follows by squaring the positive quantities and comparing \(8>1\). The homotopy is $$ H(t,z)=(1+i)z^2+(1-t)\bigl((1-i)z+1\bigr). $$ For all \(z\in C_2\) and \(t\in[0,1]\), $$ |H(t,z)|\geq4\sqrt{2}-(1-t)(2\sqrt{2}+1) \geq4\sqrt{2}-(2\sqrt{2}+1)=2\sqrt{2}-1>0. $$ At \(t=0\) the original polynomial is recovered, and at \(t=1\) the value is \((1+i)z^2\). Check: the coefficient moduli used above are \(|1+i|=\sqrt2\) and \(|1-i|=\sqrt2\), so the estimate applies to complex values without treating the coefficients as real.
Worked Example: Choosing a Radius for a Cubic
Consider \(p(z)=z^3+2z^2+3z+1\). We can verify a convenient radius directly: choose \(R=4\). The leading-term bound is $$ |z^3|=4^3=64, $$ and the sum of the lower-term bounds is $$ 2\cdot4^2+3\cdot4+1=32+12+1=45. $$ Thus the dominance gap is \(64-45=19>0\). The interpolation is $$ H(t,z)=z^3+(1-t)(2z^2+3z+1). $$ For any \(z\in C_4\) and \(t\in[0,1]\), $$ |H(t,z)|\geq64-(1-t)45\geq64-45=19>0. $$ The endpoints are \(H(0,z)=z^3+2z^2+3z+1\) and \(H(1,z)=z^3\). Check: substituting the selected radius into the leading bound gives \(4^3=64\), and substituting it into the lower bound gives \(32+12+1=45\); the strict inequality required by the theorem is \(64>45\).
What the Estimate Does—and Does Not—Say
The homotopy is a straight-line interpolation in the complex plane, performed separately at each point of \(C_R\). The parameter \(t\) changes the lower-degree contribution from its full value to zero, while leaving the leading term fixed. The triangle inequality guarantees that this cancellation cannot remove the leading contribution when the leading modulus exceeds the total lower-term bound.
Common pitfall. It is incorrect to infer that the straight-line homotopy avoids zero merely because both endpoint maps avoid zero. For a concrete counterexample, let \(p(z)=z-2\) on \(C_1\). On that circle, \(z\ne0\), and \(z-2\ne0\), because \(z=2\) would have modulus \(2\), not \(1\). Nevertheless the interpolation from \(p\) to its leading monomial \(z\) is $$ H(t,z)=(1-t)(z-2)+tz=z-2(1-t). $$ At \(t=\tfrac12\) and \(z=1\in C_1\), direct substitution gives \(H(\tfrac12,1)=1-2(1-\tfrac12)=1-1=0\). The dominance condition fails here: the leading bound is \(1\), while the lower-term bound is \(2\). This example shows why the theorem uses a radius with a strict estimate rather than only checking the endpoints.
Once a radius satisfies strict dominance, the theorem concerns the restrictions of the two polynomial expressions to that circle. It does not assert that the same interpolation is nonzero throughout the disk, or at every point of the complex plane. Its conclusion is precisely a homotopy on \(C_R\) whose values lie in \(\mathbb{C}\setminus\{0\}\). The strict inequality is what supplies the positive margin needed for that conclusion.
Check Your Understanding
For each question, distinguish the pointwise estimate on the circle from the conclusion about the whole homotopy.
- For \(p(z)=3z^2-z+2\), test whether \(R=2\) satisfies strict dominance, and write the corresponding interpolation.
- Explain why a zero coefficient among the lower-degree terms causes no difficulty in the radius-existence proof.
- For a nonzero constant polynomial \(p(z)=c\), state the leading monomial and describe the homotopy to it.
- In the proof of the theorem, identify the inequality that gives a bound uniform in both \(t\in[0,1]\) and \(z\in C_R\).
- For \(p(z)=z-3\) on \(C_1\), check the dominance inequality and determine whether the theorem guarantees that the straight-line interpolation to \(z\) is zero-free.