Factorisation over the complex numbers | Year 12

01 / LEARN

The idea, step by step

The coefficient field matters

A polynomial such as $z^{2} + 1$ has no real roots, so is irreducible into real linear factors. Over the complex numbers it is $(z - i) (z + i)$ . State the number system when a question asks for complete factorisation.

Every degree-n nonconstant complex polynomial has n roots counting multiplicity, so it factors into n linear factors over the complex numbers. This result guarantees a factorisation exists; finding its exact roots can still require suitable algebra or technology.

Use structure before expansion

Treat a polynomial in $z^{2}$ as a quadratic in a new variable, or use a known root and polynomial division. Once a root $a$ is found, the factor theorem gives the factor $z - a$ . Include a non-unit leading coefficient in the final factorisation.

For $z^{n} - a^{n}$ , polar roots can provide a direct complete factorisation. Repeated roots produce repeated factors; listing distinct values alone does not show their multiplicities.

A common trap

Stopping at irreducible real quadratics when complete factorisation over $ℂ$ is requested.

Instead: Over $ℂ$ , those quadratics split into linear factors using their non-real roots.

02 / WORKED EXAMPLES

See the thinking

Example 1

A real quadratic splits over the complex numbers

Factorise $z^{2} - 4 z + 13$ over $ℂ$ .

The quadratic formula gives $z = (4 \pm \sqrt{- 36}) / 2 = 2 \pm 3 i$ .
Each root a contributes the factor $z - a$ .
The factors multiply to ${(z - 2)}^{2} + 9 = z^{2} - 4 z + 13$ .

(z - (2 + 3 i)) (z - (2 - 3 i))

Example 2

A polynomial in a square

Factorise $z^{4} + 5 z^{2} + 4$ completely over $ℂ$ .

Put $u = z^{2}$ ; $u^{2} + 5 u + 4 = (u + 1) (u + 4)$ .
Thus the real quadratic factors are $(z^{2} + 1) (z^{2} + 4)$ .
Split each using roots $\pm i$ and $\pm 2 i$ .

(z - i) (z + i) (z - 2 i) (z + 2 i)

Check your understanding

Factorise $z^{2} + 9$ over $ℂ$ .

Show answer

$(z - 3 i) (z + 3 i)$ .

The roots are $\pm 3 i$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Factorise $z^{2} - 2 z + 5$ over $ℂ$ .
2 marks
Factorise $2 z^{2} + 8$ over $ℂ$ .
2 marks
Factorise $z^{4} - 1$ completely over $ℂ$ .
2 marks
Solve $z^{4} + 13 z^{2} + 36 = 0$ .
3 marks
Give the roots and multiplicities of ${(z^{2} + 1)}^{2} (z - 2)$ .
3 marks
Why do $z^{2} + 1$ and $z^{2} - 1$ have different real factorisations but both split over $ℂ$ ?
3 marks

Read the worked solutions

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

Factorise $z^{2} - 2 z + 5$ over $ℂ$ .
2 marks
Complete the square: ${(z - 1)}^{2} + 4$ , with roots $1 \pm 2 i$ .
$(z - (1 + 2 i)) (z - (1 - 2 i))$ .
Factorise $2 z^{2} + 8$ over $ℂ$ .
2 marks
Take out 2, then use the roots of $z^{2} + 4 = 0$ .
$2 (z - 2 i) (z + 2 i)$ .
Factorise $z^{4} - 1$ completely over $ℂ$ .
2 marks
$z^{4} - 1 = (z^{2} - 1) (z^{2} + 1)$ ; split both quadratics.
$(z - 1) (z + 1) (z - i) (z + i)$ .
Solve $z^{4} + 13 z^{2} + 36 = 0$ .
3 marks
Treat as a quadratic in $z^{2}$ : $(z^{2} + 4) (z^{2} + 9) = 0$ .
$z = \pm 2 i, \pm 3 i$ .
Give the roots and multiplicities of ${(z^{2} + 1)}^{2} (z - 2)$ .
3 marks
$z^{2} + 1 = (z - i) (z + i)$ ; the outer square doubles both multiplicities.
$i$ and $- i$ each have multiplicity 2; 2 has multiplicity 1.
Why do $z^{2} + 1$ and $z^{2} - 1$ have different real factorisations but both split over $ℂ$ ?
3 marks
The available linear factors depend on which coefficient field is allowed.
The first has non-real roots $\pm i$ ; the second has real roots $\pm 1$ . All four values are complex numbers.

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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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