Loav · VCE Specialist Mathematics · Units 3 & 4
S34-C03 · Algebra, number and structure · Complex numbers
1. What changes when roots may be complex?
A polynomial equation may have no real solution and still have complex solutions. For example, z² + 9 = 0 gives z = ±3i. The algebraic tools remain familiar: factorisation, completing the square, the remainder theorem and polynomial division all work over the complex numbers. What changes is the range of permissible roots and therefore how far a polynomial can be factored.
A root α of p is a complex number satisfying p(α) = 0. The factor theorem says that this is equivalent to z − α being a factor of p(z). A root is a number; a factor is an expression. If a question asks for factorisation, a bare list of roots is not the requested final form. If it asks for all solutions, finish by listing the values rather than leaving only a product equal to zero.
Keep the leading coefficient. If p(z) has leading coefficient a and roots α₁ through αn, counted with multiplicity, its complete linear factorisation is a(z − α₁)⋯(z − αn). Omitting a changes the polynomial even though it leaves the zero set unchanged. Exact substitution or re-expansion checks both the roots and the coefficient.
2. Complete the square before reaching for a general formula
For a monic quadratic z² + bz + c = 0, completing the square gives (z + b/2)² = b²/4 − c. This identity is valid whether b and c are real or complex. If the right side is a negative real number −d with d > 0, its two square roots are ±i√d. Both signs must be carried through to the solution for z.
For a nonzero general complex right side, find its two square roots using polar form or by setting the root equal to u + vi and comparing parts. The equation (u + vi)² = A + Bi gives u² − v² = A and 2uv = B, with u and v real. The two square roots are negatives of each other. If the right side is zero, the two formal signs coincide and there is one distinct root with multiplicity two.
The quadratic formula is another expression of this completed-square argument. For az² + bz + c = 0 with a ≠ 0, z = [−b ± √(b² − 4ac)]/(2a), where either square root of the discriminant generates the same pair through ±. For real coefficients, a positive discriminant gives two distinct real roots, zero gives a repeated real root, and a negative discriminant gives a non-real conjugate pair.
Worked example 1: complete a square with a complex constant
Solve z² − 2z + 1 − 2i = 0. Completing the square gives (z − 1)² = 2i. Since (1 + i)² = 1 + 2i + i² = 2i, the two square roots of 2i are 1 + i and −1 − i. Therefore z − 1 = ±(1 + i), giving z = 2 + i or z = −i.
The factorisation is [z − (2 + i)](z + i). Expanding gives z² + iz − (2 + i)z − (2 + i)i = z² − 2z + 1 − 2i, as required. These roots are not conjugates, and there is no reason they should be: the polynomial has a non-real constant coefficient. A real-looking coefficient of z does not make the entire polynomial real-coefficient.
3. Why real coefficients force conjugate pairs
For p(z) = anzn + ⋯ + a₀ with every coefficient real, conjugating p(α) gives p(). This follows because conjugation distributes over sums and products and leaves each real coefficient unchanged. If p(α) = 0, then p() = = 0. Thus the conjugate of every root is also a root.
For a non-real root α = u + vi with v ≠ 0, the conjugate u − vi is distinct. Their two linear factors combine to a real quadratic:
= (z − u)² + v² = z² − 2uz + u² + v².
This quadratic is often the most efficient starting point for division. Its real coefficients avoid carrying complex arithmetic through every line. The theorem also preserves multiplicity: a repeated non-real root of a real-coefficient polynomial has a conjugate root with the same multiplicity. Consequently, non-real roots occur in pairs when counted with multiplicity.
Do not apply the theorem to an arbitrary complex-coefficient polynomial. The proof fails because conjugation changes its coefficients. The polynomial z − i has the root i but not −i. Also, a real root is already equal to its conjugate; the theorem does not create an additional distinct root. A real-coefficient polynomial of odd degree must have at least one real root because an odd total cannot be filled entirely by non-real pairs.
Worked example 2: use one root to factor a real cubic
The polynomial p(z) = z³ − 3z² + z + 5 has the root 2 + i. Find every root and factor p over the complex numbers. All coefficients are real, so 2 − i is also a root. Their combined factor is (z − 2)² + 1 = z² − 4z + 5.
Divide p(z) by this quadratic. The first quotient term is z: subtracting z³ − 4z² + 5z leaves z² − 4z + 5. The next quotient term is 1, and subtraction leaves remainder zero. Thus p(z) = (z + 1)(z² − 4z + 5) = (z + 1)[z − (2 + i)][z − (2 − i)].
The roots are −1, 2 + i and 2 − i, each simple. The linear factors give three roots counted with multiplicity, matching the degree. As a second check, the sum is 3, which agrees with the negative coefficient of z² in this monic cubic. The conjugate theorem supplied one missing root; division supplied the remaining one.
4. Count roots with multiplicity
The fundamental theorem of algebra states that every nonconstant polynomial with complex coefficients has a complex root. Repeatedly applying the factor theorem gives a complete factorisation into linear factors over the complex numbers. A degree-n polynomial therefore has exactly n roots counted with multiplicity. This is an informal use of the theorem; a proof of the theorem itself is not needed here.
Multiplicity records repeated factors. If p(z) = (z − 2)²(z² + 4), then 2 is a root of multiplicity two, while 2i and −2i each have multiplicity one. There are three distinct roots but four roots counted with multiplicity. A root list with repeated values is less clear than stating the multiplicities explicitly. The repeated factor is part of the polynomial’s structure, even though it corresponds to a single Argand point.
The theorem concerns a nonconstant polynomial with a nonzero leading coefficient. A nonzero constant polynomial has no roots. The zero polynomial is zero for every z and does not have a finite root count. When parameters are present, check whether the nominal leading coefficient can vanish before applying a degree-based count.
Factorisation over the real numbers and over the complex numbers are different stopping points. The expression (z − 2)²(z² + 4) is fully factored into real linear and irreducible quadratic factors. Over the complex numbers, continue to (z − 2)²(z − 2i)(z + 2i). Always read which number system the question specifies.
Worked example 3: factor a complex-coefficient quadratic
Let q(z) = z² − (3 + i)z + 2 + 2i. Show that 1 + i is a root, then find the other root. Substitution gives (1 + i)² − (3 + i)(1 + i) + 2 + 2i = 2i − (2 + 4i) + 2 + 2i = 0. Therefore z − (1 + i) is a factor.
Matching the leading term and constant, or carrying out division, gives q(z) = [z − (1 + i)](z − 2). Indeed, the sum of the roots is 3 + i and their product is 2 + 2i, matching the quadratic. The other root is 2, not 1 − i. Direct substitution at 1 − i gives −2 + 2i, confirming that the conjugate is not a root. The correct method uses the known factor; it does not import a theorem whose coefficient condition fails.
5. Choose structure before brute-force calculation
Inspect the exponents and coefficients first. A polynomial such as z⁴ + 5z² + 4 is quadratic in z². Setting u = z² gives u² + 5u + 4, which factors easily. After solving for u, return to z and solve both resulting quadratic equations. Stopping at the u-values would answer the substituted problem rather than the original one.
For equations zn = w, the polar-root method from C02 may be the most efficient route to every linear factor. For a supplied non-real root of a real polynomial, form the conjugate quadratic. For a supplied root of a complex-coefficient polynomial, divide by its single linear factor. No general cubic or quartic formula is needed for these structured problems.
Technology can expand a proposed factorisation, divide polynomials or approximate roots. Ask for exact output where the problem permits it, and check that the solver is working over the complex numbers. A real-domain solver may return only the real subset. Numerical roots with tiny imaginary parts may approximate real roots; do not declare a new exact root type from rounding noise. Compare residuals and multiplicities with an exact factorisation when available.
In assessment work, write the real-coefficient condition before invoking conjugate pairs, show the division or coefficient comparison that finds remaining factors, and finish with the requested factorisation or root set. A correct pair alone may be incomplete when the degree indicates further roots. A sketch should label all distinct points, while repeated roots should be described in words beside the relevant point.
Practise, then check
Check each final factorisation by expansion or substitution, and count roots with multiplicity.
- Solve z² + 6z + 13 = 0 by completing the square.
- Factor 2z² + 8 over the complex numbers, keeping the leading coefficient.
- A real polynomial p(z) = z³ + az² + bz + 10 has a root −1 + 2i. Find real a and b and all roots.
- Solve z⁴ + 5z² + 4 = 0.
- For p(z) = (z + 3)²(z − i)³(z + i)³, state its degree, distinct roots and their multiplicities. Does p have real coefficients?
- The polynomial z² − (2 + i)z + 2i has root i. Find its other root and explain why −i need not be a root.
- A real-coefficient polynomial of degree five has roots 1 + 2i and −i, each simple. What can be concluded about its remaining roots?
Reasoned answers
- Completing the square gives (z + 3)² + 4 = 0, so z + 3 = ±2i. The roots are −3 + 2i and −3 − 2i. Both satisfy the equation, and the pair agrees with the real coefficients and negative discriminant.
- Write 2(z² + 4) = 2(z − 2i)(z + 2i). The roots are ±2i, but the factor 2 is needed to reproduce the original polynomial. Expanding the linear factors gives z² − (2i)² = z² + 4.
- The conjugate −1 − 2i is also a root, so z² + 2z + 5 is a factor. Since the cubic is monic and has constant 10, the remaining factor is z + 2. Expansion gives z³ + 4z² + 9z + 10. Thus a = 4, b = 9 and the third root is −2.
- Put u = z². Then u² + 5u + 4 = (u + 1)(u + 4), so z² = −1 or z² = −4. The four roots are i, −i, 2i and −2i. They are distinct and give the required total of four.
- The degree is 2 + 3 + 3 = 8. The distinct roots are −3 with multiplicity two, i with multiplicity three and −i with multiplicity three. Combining the conjugate factors gives (z + 3)²(z² + 1)³, which visibly has real coefficients.
- Factor as (z − i)(z − 2). The other root is 2. The coefficient of z and the constant are non-real, so the real-coefficient conjugate-root theorem does not apply. Substitution at −i gives −2 + 4i, which is nonzero.
- The conjugates 1 − 2i and i must also be roots, each simple. These four non-real roots account for four degrees, leaving one root counted with multiplicity. It must be real, because any additional non-real root would require a distinct conjugate and two remaining degrees. Its numerical value cannot be determined from the information given.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4. Scope follows the VCAA Mathematics Study Design 2023–2027, Specialist Mathematics Areas of Study 1 and 3 as applicable, and the current Specialist Mathematics FAQ. Unit 2 complex-number foundations are retained where required. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 6E–6F, pp. 262–272. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
