Conjugate-root results for real polynomials | Year 12

01 / LEARN

Understand the idea.

Why real coefficients force conjugate pairs

For a polynomial with real coefficients, conjugation commutes with evaluation: $P (\bar{z}) = \bar{P (z)}$ . This follows because conjugation preserves sums and products and leaves each real coefficient unchanged. A zero value therefore forces a zero value at the conjugate input.

Thus a non-real root $a + b i$ forces the root $a - b i$ , with the same multiplicity. The real-coefficient condition matters: $P (z) = z - i$ has root $i$ without root $- i$ . A real root already equals its own conjugate.

Build a real quadratic from a pair

Pair conjugate factors before expanding: $(z - a - b i) (z - a + b i) = {(z - a)}^{2} + b^{2}$ . The imaginary terms cancel, producing a real quadratic. This is useful for reconstructing real polynomials from known roots.

Count roots with multiplicity to check completeness. Non-real roots use an even number of places because they pair up, so a real polynomial of odd degree must have at least one real root. It can have more than one real root.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Recover a real cubic

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

Worked solution

1
Supply the forced conjugate root.
$1 + 2 i is a root ⟹ 1 - 2 i is a root$
All coefficients are real, so the conjugate-root theorem applies.
2
Multiply the conjugate factors.
$\begin{aligned} (z - 1 - 2 i) (z - 1 + 2 i) & = {(z - 1)}^{2} + 4 \\ = z^{2} - 2 z + 5 \end{aligned}$
A conjugate pair gives a real quadratic.
3
Include the remaining root.
$P (z) = (z - 3) (z^{2} - 2 z + 5)$
The three listed roots fill the degree-three polynomial, and monic means leading coefficient one.
4
Expand and collect terms.
$\begin{aligned} P (z) & = z^{3} - 2 z^{2} + 5 z - 3 z^{2} + 6 z - 15 \\ = z^{3} - 5 z^{2} + 11 z - 15 \end{aligned}$
Combine coefficients of the same power of $z$ .

Answer

Third root $1 - 2 i$ ; $P (z) = z^{3} - 5 z^{2} + 11 z - 15$ .

EXAMPLE 02

Find unknown real coefficients

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

Worked solution

1
Identify the other root.
$- 2 - 3 i$
The real coefficients force the conjugate of the given non-real root.
2
Build the monic quadratic.
$\begin{aligned} z^{2} + a z + b & = (z + 2 - 3 i) (z + 2 + 3 i) \\ = {(z + 2)}^{2} + 9 \end{aligned}$
The factor for a root $- 2 \pm 3 i$ is $z + 2 \mp 3 i$ .
3
Expand and compare coefficients.
$z^{2} + a z + b = z^{2} + 4 z + 13 ⟹ a = 4, b = 13$
Identical monic polynomials have equal corresponding coefficients.

Answer

$a = 4$ , $b = 13$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$2 - i$ .

The real coefficients force $2 - i$ to accompany $2 + i$ . Together with the real root $5$ , these account for all three roots of the cubic, counting multiplicity.

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Why real coefficients force conjugate pairs

Build a real quadratic from a pair

See the method in action.

Recover a real cubic

Supply the forced conjugate root.

Multiply the conjugate factors.

Include the remaining root.

Expand and collect terms.

Find unknown real coefficients

Identify the other root.

Build the monic quadratic.

Expand and compare coefficients.

Try it yourself.

Check your understanding

Keep practising