CAIE A-Level · Mathematics 9709 · Complex Numbers

Complex Roots of Polynomials — Conjugate Pairs (9709 P3)

9 min readSyllabus 3.9PreviewBy Uzair Khan

Syllabus objective

Use the result that, for a polynomial equation with real coefficients, any non-real roots occur in conjugate pairs, e.g. in solving a cubic or quartic equation where one complex root is given.

Introduction

When solving polynomial equations in Pure Mathematics 3, it is common to encounter roots that are complex numbers. A powerful and examinable result — the conjugate root theorem — tells us that for any polynomial with real coefficients, complex roots can never appear alone: they always come in conjugate pairs. Examiners exploit this repeatedly by giving one complex root of a cubic or quartic and asking you to find the remaining roots. Mastering this result is essential for every complex numbers question involving polynomials.


Core Concept

The Conjugate Root Theorem:

If p(x)p(x) is a polynomial with real coefficients and z=a+biz = a + bi (where b0b \neq 0) is a root, then its complex conjugate zˉ=abi\bar{z} = a - bi is also a root.

Why this matters structurally

Because complex roots come in pairs, we can immediately deduce:

  • A cubic with real coefficients and one non-real root must have one real root and one conjugate pair.
  • A quartic with real coefficients can have zero, two, or four non-real roots (always in conjugate pairs).

Forming a real quadratic factor

Given a conjugate pair z=a+biz = a + bi and zˉ=abi\bar{z} = a - bi, we can form a quadratic factor with real coefficients:

(xz)(xzˉ)=x22ax+(a2+b2)(x - z)(x - \bar{z}) = x^2 - 2ax + (a^2 + b^2)

This quadratic has real coefficients (the imaginary parts cancel), and its discriminant is negative, confirming no further real roots from this factor. This quadratic factor is the key tool for reducing a cubic or quartic to something you can solve completely.


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