LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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A nonzero value of a nonconstant complex polynomial cannot be a local minimum of its modulus

Statement

If $p$ is nonconstant and $p(a)\ne0$ , then $|p|$ is not minimal on any neighbourhood of $a$ . The conventions and prerequisite facts used below are recorded in Formal complex polynomials, evaluation, degree, leading coefficient, and monic polynomials, Conjugation laws, $z\overline z=|z|^2$ , multiplicativity of modulus, and the triangle inequality, The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n\ge1$ , The binomial theorem over the complex field, Laws of finite sums and finite products.

Facts & Assumptions

Given: A nonconstant polynomial $p$ and a point $a$ with $p(a)\ne0$ .

[L1]

The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n\ge1$ supplies an $m$ th root of every nonzero complex number when $m\ge1$ .

[L2]

Conjugation laws, $z\overline z=|z|^2$ , multiplicativity of modulus, and the triangle inequality gives $|uv|=|u||v|$ , the triangle inequality, and $v\overline v=|v|^2$ .

[L3]

The binomial theorem over the complex field gives the finite expansion of $(a+z)^n$ in complex coefficients.

Proof

technique · constructive

1.1

By [L3], expanding $p(a+z)-p(a)$ gives a nonzero polynomial in $z$ (its top coefficient is the nonzero leading coefficient of $p$ ). Let $m\ge1$ be its first nonzero degree, so $p(a+z)=p(a)+c_mz^m+z^{m+1}q(z)$ with $c_m\ne0$ . Choose by [L1] a unit complex number $u$ with $u^m=-p(a)\overline{c_m}/(|p(a)||c_m|)$ . Then $\overline{p(a)}c_mu^m=-|p(a)||c_m|$ .

L1L3constructalgebra

1.2

Write $M$ for the sum of the moduli of the finitely many coefficients of $q$ . For $0<t\le1$ , [L2] gives $|t^{m+1}q(tu)|\le Mt^{m+1}$ . With $\lambda:=|p(a)||c_m|$ and $B:=2|p(a)|M+(|c_m|+M)^2>0$ , choose $0<t<\min\{1,\lambda/B\}$ .

L2choose

2.1

Put $R=t^{m+1}q(tu)$ . By step 1.1 and [L2], $|p(a+tu)|^2\le |p(a)|^2-2\lambda t^m+2|p(a)|Mt^{m+1}+(|c_m|+M)^2t^{2m}\le |p(a)|^2-2\lambda t^m+Bt^{m+1}<|p(a)|^2$ . Thus $a+tu$ is arbitrarily close to $a$ and has strictly smaller modulus.

L2step 1.1step 1.2discharge-construct∎

Depends on

Used by

Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 103 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I: The Fundamental Theorem of Algebra (standard reference, not scraped)