Factorisation over the complex numbers | Year 12

01 / LEARN

Understand the idea.

The coefficient field matters

The requested coefficient field determines when a factorisation is complete. For example, $z^{2} + 1$ has no real linear factors, but over $ℂ$ it splits as $(z - i) (z + i)$ . A non-real root is just as valid there as a real root.

The fundamental theorem of algebra guarantees that a degree- $n$ polynomial over $ℂ$ has $n$ roots counting multiplicity. Its complete factorisation consists of a leading coefficient and $n$ linear factors, including repeated factors when a root is repeated.

Use structure before expansion

Look for algebraic structure before solving. A polynomial containing only even powers may be quadratic in $z^{2}$ ; substitute a temporary variable, factor it, and then return to $z$ . Each remaining quadratic must also split when working over $ℂ$ .

The factor theorem links a root $a$ to the factor $z - a$ . Keep the sign inside that factor correct, retain any leading coefficient, and expand the final product to check that it recovers the original polynomial.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A real quadratic splits over the complex numbers

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

Worked solution

1
Calculate the discriminant.
$Δ = {(- 4)}^{2} - 4 (1) (13) = - 36$
A negative real discriminant leads to two non-real quadratic roots.
2
Apply the quadratic formula.
$z = \frac{4 \pm 6 i}{2} = 2 \pm 3 i$
The two square roots of $- 36$ are $\pm 6 i$ .
3
Form the linear factors.
$z^{2} - 4 z + 13 = (z - (2 + 3 i)) (z - (2 - 3 i))$
Each root $a$ contributes $z - a$ , and the original leading coefficient is one.
4
Check the conjugate product.
$(z - 2 - 3 i) (z - 2 + 3 i) = {(z - 2)}^{2} + 9$
The product expands to $z^{2} - 4 z + 13$ , confirming the signs and constant.

Answer

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

EXAMPLE 02

A polynomial in a square

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

Worked solution

1
Treat the expression as a quadratic.
$u = z^{2}, u^{2} + 5 u + 4 = (u + 1) (u + 4)$
The factors have product $4$ and their constants sum to $5$ .
2
Restore the original variable.
$z^{4} + 5 z^{2} + 4 = (z^{2} + 1) (z^{2} + 4)$
Substituting $u = z^{2}$ preserves the polynomial identity.
3
Split each quadratic over the complex numbers.
$z^{2} + 1 = (z - i) (z + i), z^{2} + 4 = (z - 2 i) (z + 2 i)$
The roots are $\pm i$ and $\pm 2 i$ .
4
Combine every linear factor.
$z^{4} + 5 z^{2} + 4 = (z - i) (z + i) (z - 2 i) (z + 2 i)$
Four linear factors account for the degree-four polynomial.

Answer

$(z - i) (z + i) (z - 2 i) (z + 2 i)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Factorise $z^{2} + 9$ over $ℂ$ .

Show answer & reasoning

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

The equation $z^{2} = - 9$ has roots $z = \pm 3 i$ , so the factors are $(z - 3 i) (z + 3 i)$ . Their product is $z^{2} - {(3 i)}^{2} = z^{2} + 9$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

The coefficient field matters

Use structure before expansion

See the method in action.

A real quadratic splits over the complex numbers

Calculate the discriminant.

Apply the quadratic formula.

Form the linear factors.

Check the conjugate product.

A polynomial in a square

Treat the expression as a quadratic.

Restore the original variable.

Split each quadratic over the complex numbers.

Combine every linear factor.

Try it yourself.

Check your understanding

Keep practising