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AS Level · Further Pure Mathematics 1 WFM01

Solving Equations with Complex Roots

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Edexcel IAL Further Pure 1

Solving Equations
with Complex Roots

When a polynomial's discriminant turns negative, real numbers aren't enough — but complex numbers always are. Every polynomial equation has exactly as many roots as its degree, once you allow complex values.

Quadratics · Cubics · Quartics Complex Conjugate Pairs ~25 min read

The Big Idea

Whenever a polynomial with real coefficients has complex roots, those roots always come in conjugate pairs — if a + bi is a root, then so is a − bi. This one rule unlocks every complex-root problem in this chapter.

Chapter Overview

A quadratic with a negative discriminant (b² − 4ac < 0) has two complex conjugate roots.
Complex roots always appear as conjugate pairs (a + bi and a − bi) when all coefficients are real.
You can build a quadratic from one complex root using (z − z₁)(z − z₁*).
A cubic with real coefficients has either 3 real roots or 1 real + 1 conjugate pair.
A quartic with real coefficients has either 4 real, 2 real + 1 pair, or 2 pairs of complex roots.
To solve higher-degree equations, extract the known quadratic factor and divide/equate coefficients for the remainder.
The conjugate pair gives a quadratic factor; this is more efficient than polynomial long division when coefficients are algebraic.

Topic 1

Quadratics with Complex Roots

1.1 — Why does the quadratic formula give complex roots?

The quadratic formula is z = (−b ± √(b²−4ac)) / 2a. When the discriminant b² − 4ac is negative, you're taking the square root of a negative number — which is impossible in the real numbers but perfectly fine in the complex numbers.

The trick is to rewrite it using i, where i² = −1:

Key Rewriting Rule

√(−k) = √k · i (for k > 0)

e.g. √(−144) = √144 · i = 12i | √(−2) = √2 · i

Once you rewrite the square root using i, the quadratic formula proceeds as normal and gives you two complex answers.

Why are the two roots conjugates? Look at the "±" in the formula: one root gets +√(−k) and the other gets −√(−k). That's exactly what makes them conjugates — they share the same real part and have imaginary parts that are equal in size but opposite in sign.

The Two Complex Roots of az² + bz + c = 0

z₁ = x + iy and z₂ = x − iy

They are always complex conjugates of each other: z₂ = z₁*

Worked Example — Solving a quadratic with complex roots

Solve z² − 2z + 37 = 0.

Identify a = 1, b = −2, c = 37. Calculate the discriminant:

b² − 4ac = (−2)² − 4(1)(37) = 4 − 148 = −144

Negative discriminant confirms complex roots.

Apply the quadratic formula:

z = (−(−2) ± √(−144)) / (2 × 1) = (2 ± √(−144)) / 2

Rewrite √(−144) = 12i:

z = (2 ± 12i) / 2

Check: they are complex conjugates of each other — always verify this.

Practice Question

Solve z² + 4z + 13 = 0.

1.2 — One root known: instantly write the other

If the equation has real coefficients and you're told one complex root, the other root is automatically its complex conjugate. No solving required.

The shortcut: If z₁ = a + bi is a root of a polynomial with real coefficients, then z₂ = a − bi is also a root. Just flip the sign of the imaginary part.

4 + 5i z² − 8z + 41 = 0 4 − 5i

Topic 2

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Also in the full note

Cubics & Quartics with Complex Roots
What to Memorise
Concepts Checklist
Exam Tips & Common Mistakes
Formula Summary Card
1.3 — Building a quadratic from a complex root
2.1 — How many real vs complex roots are possible?
2.2 — Solving a cubic with one known complex root

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Solving Equations with Complex Roots

Solving Equationswith Complex Roots

Quadratics with Complex Roots

1.1 — Why does the quadratic formula give complex roots?

1.2 — One root known: instantly write the other

More Further Pure Mathematics 1 topics

Solving Equations
with Complex Roots