Dominance Estimates for Polynomials on Large Circles

Why a Large-Circle Estimate Is Useful

A polynomial is a sum of terms with different powers of its input. When the input has large modulus, its highest-degree term grows faster than every lower-degree term. To use this fact rigorously, one needs a radius at which the leading term exceeds the sum of the magnitudes of all the other terms. That strict inequality is a uniform estimate: it holds at every point on the circle, regardless of the argument of the complex input.

Throughout, \(\mathbb{C}\) has its usual modulus \(|z|\), and a circle of radius \(R>0\) means the set of all \(z\in\mathbb{C}\) with \(|z|=R\). Consider a nonconstant polynomial \(P(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0\), where \(n\ge1\), \(a_n\ne0\), and the coefficients \(a_k\) are complex numbers. Some lower coefficients may be zero.

The second condition is useful because the triangle inequality gives \(|L(z)|\le\sum_{k=0}^{n-1}|a_kz^k|\). It is sufficient, not necessary: different lower terms can cancel, so their sum can have smaller modulus than the sum of their individual moduli.

An Explicit Radius Bound

Normalize the lower coefficients by the size of the leading coefficient, and set \(M=\sum_{k=0}^{n-1}|a_k|/|a_n|\). This number is finite and nonnegative. The choice \(R=1+M\) is a convenient explicit bound; it is not intended to be the smallest possible radius.

Proof. Fix \(z\) with \(r=|z|\ge R=1+M\). Then \(r\ge1\). For each integer \(k\) with \(0\le k\le n-1\), we have \(r^k\le r^{n-1}\), including \(k=0\), since \(r\ge1\). Therefore \(\sum_{k=0}^{n-1}|a_kz^k|=\sum_{k=0}^{n-1}|a_k|r^k\le r^{n-1}\sum_{k=0}^{n-1}|a_k|\). The definition of \(M\) gives \(\sum_{k=0}^{n-1}|a_k|=|a_n|M\), so the last expression is \(|a_n|Mr^{n-1}\). Since \(r\ge1+M>M\), and \(|a_n|r^{n-1}>0\), we have \(|a_n|Mr^{n-1}<|a_n|r^nr^0=|a_n|r^n=|a_nz^n|\). This proves the claimed strict inequality. If \(M=0\), the same reasoning still applies: the lower-term sum is zero, while \(|a_nz^n|>0\) because \(r\ge1\) and \(a_n\ne0\). Thus the degenerate case of a polynomial with no nonzero lower terms is included. \(\square\)

The proof uses the modulus estimate \(r^k\le r^{n-1}\), which is why the bound includes the condition \(r\ge1\). It also explains why the conclusion holds not only on the chosen circle but at every point outside it. No cancellation or choice of angle is needed.

Proof. Write \(P(z)=a_nz^n+L(z)\), where \(L(z)=\sum_{k=0}^{n-1}a_kz^k\). The triangle inequality applied to \(a_nz^n=P(z)-L(z)\) gives \(|a_nz^n|\le|P(z)|+|L(z)|\), and hence \(|P(z)|\ge|a_nz^n|-|L(z)|\). Also, \(|L(z)|\le\sum_{k=0}^{n-1}|a_kz^k|\). The theorem makes this final sum strictly less than \(|a_nz^n|\), so \(|P(z)|>0\). \(\square\)

Worked Estimates

What the Bound Does—and Does Not—Say

The radius \(1+M\) is a safe choice obtained by replacing every lower power \(r^k\) by the largest lower power \(r^{n-1}\). This can overestimate the lower-term sum, especially when many coefficients vanish or when most nonzero coefficients occur at low degrees. A smaller radius may still work; the theorem gives a bound that is easy to compute and proves the desired inequality without solving an optimization problem.

There is also a sharper pointwise inequality available before choosing a radius. For \(r=|z|\), the ratio of the lower-term sum to the leading-term modulus is bounded by \(\sum_{k=0}^{n-1}(|a_k|/|a_n|)r^{k-n}\), whenever \(r>0\). Each exponent \(k-n\) is negative, so each nonzero summand decreases as \(r\) increases. The explicit theorem avoids having to solve for where this sum drops below \(1\); it uses the simpler estimate \(r^k\le r^{n-1}\) for \(r\ge1\).

Common pitfall: checking only the next-highest term. For \(Q(z)=z^3-10z^2+100z+1\), one might incorrectly focus on the comparison \(r^3>10r^2\), conclude that \(r>10\) suffices, and ignore the remaining terms. At \(r=11\), the leading modulus is \(11^3=1331\), but the sum of lower-term moduli is \(10\cdot11^2+100\cdot11+1=10\cdot121+1100+1=1210+1100+1=2311\). Thus the required dominance fails at radius \(11\). The theorem uses every lower coefficient: \(M=(10+100+1)/1=111\), so \(R=112\). At that radius, the leading modulus is \(112^3=1404928\), while the lower-term sum is \(10\cdot112^2+100\cdot112+1=10\cdot12544+11200+1=125440+11200+1=136641\). Therefore \(1404928>136641\). The correction is not to compare the leading term with one lower term, but to bound the sum of all lower-term moduli.

This estimate is useful whenever an argument needs a circle on which the leading term controls the polynomial uniformly. Its content is strictly quantitative: it supplies a radius, proves a strict inequality at every point of that circle, and even gives the same control throughout the exterior. It does not assert that the chosen radius is optimal.