Conjugate-root results for real polynomials | Year 12

01 / LEARN

The idea, step by step

Why real coefficients force conjugate pairs

Conjugation preserves addition and multiplication. If $P (z) = a_{n} z^{n} + \dots + a_{0}$ has real coefficients, then $P (\bar{z}) = \bar{P (z)}$ because each $\bar{a_{j}} = a_{j}$ . Therefore $P (α) = 0$ implies $P (\bar{α}) = 0$ .

The real-coefficient condition is essential. For example, $P (z) = z - i$ has root i but not -i; its constant coefficient is not real. Real roots are already their own conjugates.

Build a real quadratic from a pair

If $a + b i$ and $a - b i$ are roots, with real a and nonzero b, their product of factors is $(z - a - b i) (z - a + b i) = {(z - a)}^{2} + b^{2} = z^{2} - 2 a z + (a^{2} + b^{2})$ . It has real coefficients.

The conjugate roots also have equal multiplicities for real polynomials. A real polynomial of odd degree has at least one real root because its non-real roots can be paired, counting multiplicity. This does not mean an odd-degree polynomial has only one real root.

A common trap

Applying the conjugate-root theorem to any polynomial, regardless of its coefficients.

Instead: The standard result requires every coefficient to be real; a polynomial with non-real coefficients need not have paired roots.

02 / WORKED EXAMPLES

See the thinking

Example 1

Recover a real cubic

A monic real cubic has roots $1 + 2 i$ and 3. Find its third root and expand it.

The conjugate-root result forces $1 - 2 i$ .
The conjugate factors give $z^{2} - 2 z + 5$ .
Multiply $(z - 3) (z^{2} - 2 z + 5)$ and collect terms.

Third root

1 - 2 i

;

P (z) = z^{3} - 5 z^{2} + 11 z - 15

Example 2

Find unknown real coefficients

$z^{2} + a z + b$ has real coefficients and root $- 2 + 3 i$ . Find a and b.

The other root is $- 2 - 3 i$ .
The factor product is ${(z + 2)}^{2} + 9 = z^{2} + 4 z + 13$ .
Compare coefficients with the given monic quadratic.

a = 4

b = 13

Check your understanding

A real cubic has roots $2 + i$ and 5. What is its remaining root?

Show answer

$2 - i$ .

The non-real root forces its conjugate, and all three roots are then accounted for.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find a real monic quadratic with roots $3 \pm i$ .
2 marks
A monic real quartic has roots $i$ , $2 + i$ and their conjugates. Factorise it over $ℝ$ .
2 marks
A monic real cubic has root $1 + i$ and constant term $- 6$ . Find the remaining real root.
3 marks
Explain why a real polynomial of degree 5 cannot have exactly one non-real root counting multiplicity.
3 marks
For real a and b, $z^{2} + a z + b$ has root $1 - \sqrt{2} i$ . Find a and b.
2 marks
Test the claim: if i is a root of $P (z) = z - i$ , then -i is also a root.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find a real monic quadratic with roots $3 \pm i$ .
2 marks
${(z - 3)}^{2} + 1 = z^{2} - 6 z + 10$ .
$z^{2} - 6 z + 10$ .
A monic real quartic has roots $i$ , $2 + i$ and their conjugates. Factorise it over $ℝ$ .
2 marks
The conjugate pairs are $\pm i$ and $2 \pm i$ .
$(z^{2} + 1) (z^{2} - 4 z + 5)$ .
A monic real cubic has root $1 + i$ and constant term $- 6$ . Find the remaining real root.
3 marks
The conjugate pair gives $z^{2} - 2 z + 2$ . Write $(z - r) (z^{2} - 2 z + 2)$ ; its constant term is $- 2 r = - 6$ .
$3$ .
Explain why a real polynomial of degree 5 cannot have exactly one non-real root counting multiplicity.
3 marks
One non-real root would leave its forced conjugate uncounted.
Every non-real root must be accompanied by its distinct conjugate, so the non-real count is even.
For real a and b, $z^{2} + a z + b$ has root $1 - \sqrt{2} i$ . Find a and b.
2 marks
The pair gives ${(z - 1)}^{2} + 2 = z^{2} - 2 z + 3$ .
$a = - 2$ , $b = 3$ .
Test the claim: if i is a root of $P (z) = z - i$ , then -i is also a root.
3 marks
The coefficient $- i$ is non-real, so the conjugate-root theorem does not apply.
The claim is false: $P (- i) = - 2 i \neq 0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

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