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Complex Roots of Polynomial Equations with Real Coefficients | 实系数多项式方程的复根

2026-08-21 21:33 · 本文部分内容由AI辅助创作

📚 Complex Roots of Polynomial Equations with Real Coefficients | 实系数多项式方程的复根

When you solve a polynomial equation with real coefficients, any non-real complex roots always occur in conjugate pairs. This fact is a direct consequence of the way complex conjugation interacts with addition and multiplication, and it gives a powerful shortcut in factorising, curve sketching and coefficient problems. In AQA A-Level Mathematics and Further Mathematics, questions often ask you to use one given complex root to find all roots, construct a real polynomial, or determine unknown real coefficients.

当你解一个实系数多项式方程时，任何非实复根总是以共轭对的形式出现。这一事实是复数共轭与加法、乘法相互作用的直接结果，它为因式分解、图像草图以及系数问题提供了强有力的捷径。在 AQA A-Level 数学与进阶数学中，题目经常要求你利用一个已知复根找出全部根、构造实多项式，或确定未知实系数。

This article restates the conjugate root theorem, proves it from first principles, then works through every standard exam application: counting roots, building quadratic factors, constructing polynomials, finding unknown coefficients, handling repeated roots, sketching graphs, and the trigonometric spin-offs such as the roots of unity.

本文将重述共轭根定理，从基本原理出发给出证明，然后逐一演练所有标准考点应用：根的计数、构造二次因式、构造多项式、求未知系数、处理重根、绘制图像，以及单位根等三角推论。

1. Real Coefficients and the Conjugate Root Theorem | 实系数与共轭根定理

Let p(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ be a polynomial whose coefficients aₙ, aₙ₋₁, … , then the complex conjugate α* is also a root. This is called the conjugate root theorem.

设 p(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀ 是一个多项式，其系数 aₙ, aₙ₋₁, …, a₀ 全为实数，且 aₙ ≠ 0。如果 α 是方程 p(z) = 0 的一个非实根，即 p(α) = 0，那么它的共轭复数 α* 也是根。这称为共轭根定理。

The word "non-real" is essential. A real root equals its own conjugate, so pairing tells you nothing new about it. The theorem says that genuinely complex roots arrive in pairs {α, α*} with α ≠ α*, so a real polynomial cannot have exactly one non-real root.

"非实"这个词很关键。实根等于它自己的共轭，所以配对对实根本身没有新信息。定理说的是真正的复根以 {α, α*} 成对出现，且 α ≠ α*，因此实系数多项式不可能恰好只有一个非实根。

2. Why Conjugation Works: The Proof | 共轭为何成立：证明

ates. Since every real coefficient equals its own conjugate, applying conjugation to p(α) = 0 gives p(α*) = 0.

证明利用两个性质：和的共轭等于共轭的和，积的共轭等于共轭的积。由于每个实数系数都等于它自己的共轭，对 p(α) = 0 两边取共轭就得到 p(α*) = 0。

for every positive integer k, and the constant term satisfies a₀* = a₀ because a₀ is real.

把这两条法则写清楚。对任意复数 u 与 v，有 (u + v)* = u* + v* 以及 (uv)* = u*v*。反复使用积的法则可得 (αᵏ)* = (α*)ᵏ，对任意正整数 k 成立；而常数项满足 a₀* = a₀，因为 a₀ 是实数。

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aₖ, so the conjugate of the polynomial value is the polynomial evaluated at the conjugate:

把整个方程逐项取共轭。由于每个 aₖ 都是实数，aₖ* = aₖ，所以多项式值的共轭等于在共轭点处求值的多项式：

[p(α)]* = p(α*)

at α* is a root. Notice the elegant consequence: conjugation commutes with evaluating a real polynomial.

由于 p(α) = 0 且 0* = 0，我们得到 p(α*) = 0，这正是 α* 是根这一结论。注意由此得到的漂亮推论：共轭运算与实系数多项式的求值可以交换次序。

3. Degree, Multiplicity and Counting Roots | 次数、重数与根的计数

th complex coefficients has exactly n roots when counted with multiplicity. Since the non-real roots of a real polynomial pair up, the number of non-real roots is always even. Everything about how many real roots you can have follows from that single parity observation.

代数基本定理指出，次数 n ≥ 1 的复系数多项式在计重数的意义下恰有 n 个根。由于实系数多项式的非实根成对出现，非实根的个数总是偶数。关于实根个数的所有结论都来自这一条奇偶性观察。

Degree n | 次数 nPossible real roots | 可能的实根数Possible non-real roots | 可能的非实根数
3 (odd | 奇)3, 10, 2
4 (even | 偶)4, 2, 00, 2, 4
5 (odd | 奇)5, 3, 10, 2, 4

4. From One Root to All Roots: The Quadratic Factor Method | 从一个根到全部根：二次因式法

oth (z − α) and (z − α*) are factors of p(z). Multiplying them produces a real quadratic factor that you can actually use.

已知 α = a + bi（b ≠ 0）是一个根，那么 (z − α) 与 (z − α*) 都是 p(z) 的因式。把它们相乘可以得到一个真正可用的实系数二次因式。

(z − α)(z − α*) = z² − (α + α*)z + αα* = z² − 2az + (a² + b²)

ch is strictly negative, so this quadratic is irreducible over the real numbers and cannot be factorised into two real linear factors.

和 α + α* = 2a 与积 αα* = a² + b² 都是实数，所以这个二次式是实系数。它的判别式为 Δ = (−2a)² − 4(a² + b²) = −4b²，严格小于零，因此该二次式在实数范围内不可约，无法分解成两个实一次因式。