Complex Conjugates and Polynomial Roots

The complex conjugate is the single most useful companion to a complex number. Flipping the sign of the imaginary part gives a partner $\bar z$ whose interaction with $z$ is always real — and this one fact powers division, modulus calculations, and the cornerstone result of this lesson: the conjugate root theorem, which forces complex roots of real-coefficient polynomials to appear in pairs. That theorem turns a single complex root into a complete factorisation of a cubic or quartic, which is why "given that $z$ is a root, find the others" is one of the most reliable styles of question in the whole strand — and one you should aim to answer almost mechanically.

Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2, and one of the most heavily examined parts of the complex-numbers strand. It blends AO1 (find a conjugate, form a quadratic factor, divide a polynomial) with substantial AO2 (state and use the conjugate root theorem, justify why the resulting quadratic factor is real) and AO3 (assemble these into a multi-step "find all roots" problem). It links forward to the algebra of polynomial roots (sums and products of roots) in the further-algebra strand.

Core theory: the conjugate and its algebra

For $z = a + bi$ , the complex conjugate is

$\bar z = a - bi$

— the reflection of $z$ in the real axis (geometrically; see Argand diagrams). The conjugate behaves beautifully under every operation:

Property	Formula
Involution	$\overline{\bar z} = z$
Sum / difference	$\overline{z \pm w} = \bar z \pm \bar w$
Product	$\overline{zw} = \bar z\,\bar w$
Quotient	$\overline{\left(z/w\right)} = \bar z / \bar w$
Real part	$z + \bar z = 2\operatorname{Re}(z)$
Imaginary part	$z - \bar z = 2i\operatorname{Im}(z)$
Conjugate product	$z\bar z = a^2 + b^2 =
Fixed points	$z = \bar z \iff z \in \mathbb{R}$

The two structural rules $\overline{z + w} = \bar z + \bar w$ and $\overline{zw} = \bar z\,\bar w$ say that conjugation respects all polynomial arithmetic — which is exactly what the conjugate root theorem needs. They are easy to prove directly; for the product, write $z = a + bi$ , $w = c + di$ :

$\begin{aligned} \overline{zw} &= \overline{(ac - bd) + (ad + bc)i} = (ac - bd) - (ad + bc)i, \\ \bar z\,\bar w &= (a - bi)(c - di) = (ac - bd) - (ad + bc)i, \end{aligned}$

and the two agree. The sum rule is even more immediate. By induction these extend to any finite sum or product, so $\overline{z^k} = \bar z^{\,k}$ and conjugation passes through any polynomial with real coefficients — the engine of the theorem below.

The conjugate root theorem

Theorem. If $p(x)$ is a polynomial with real coefficients and $z = a + bi$ (with $b \neq 0$ ) is a root, then $\bar z = a - bi$ is also a root.

Proof. Write $p(x) = c_n x^n + \cdots + c_1 x + c_0$ with every $c_k \in \mathbb{R}$ , so $\bar{c_k} = c_k$ . Suppose $p(z) = 0$ . Take the conjugate of both sides and use that conjugation distributes over sums and products:

$\overline{p(z)} = \sum_{k=0}^{n} \overline{c_k z^k} = \sum_{k=0}^{n} c_k \bar z^{\,k} = p(\bar z).$

But $\overline{p(z)} = \bar 0 = 0$ , so $p(\bar z) = 0$ : $\bar z$ is a root. $\blacksquare$

Consequences. Complex roots of real polynomials occur in conjugate pairs, so a real polynomial has an even number of non-real roots. Several useful facts follow immediately:

An odd-degree real polynomial has at least one real root (the non-real roots pair up, so an odd total forces at least one real). This is why every real cubic crosses the $x$ -axis at least once.
A real polynomial of degree $n$ has between $0$ and $\lfloor n/2 \rfloor$ conjugate pairs.
Every real polynomial factorises over $\mathbb{R}$ into linear and irreducible-quadratic factors only (each conjugate pair fuses into one real quadratic). There are no irreducible real polynomials of degree $3$ or higher — a fact you rely on, perhaps without realising, every time you set up partial fractions.

The theorem is also the reason these questions are answerable in an exam: a single complex root is "worth two," instantly handing you a real quadratic factor and collapsing the problem by two degrees.

The real quadratic factor

A conjugate pair $a \pm bi$ gives a quadratic factor with real coefficients:

$(x - (a+bi))(x - (a-bi)) = (x-a)^2 + b^2 = x^2 - 2ax + (a^2 + b^2).$

Equivalently, with $s = z + \bar z = 2a$ (sum) and $p = z\bar z = a^2 + b^2$ (product), the factor is $x^2 - sx + p$ . This is the bridge from "I know one complex root" to "I have a real factor I can divide out."

The standard solving strategy

Almost every "find all the roots" question follows the same four-step recipe, and writing the steps explicitly is what earns the method marks:

State the conjugate. Given a non-real root $z$ of a real polynomial, write down that $\bar z$ is also a root.
Form the real quadratic factor $x^2 - (z + \bar z)x + z\bar z$ from the pair (use sum and product — it avoids handling $i$ inside a product).
Divide the quadratic factor into the polynomial (by algebraic long division, or by comparing coefficients of an assumed quotient).
Solve the remaining factor for the last root(s).

For a cubic, step 3 leaves a linear factor and step 4 is a single division; for a quartic, the remaining factor is a quadratic which may itself have real or a second conjugate pair of roots.

Worked examples (with mark scheme)

Example 1 — conjugate identities

Given $z = 4 - 3i$ , find $z + \bar z$ , $z - \bar z$ and $z\bar z$ .

With $\bar z = 4 + 3i$ :

$z + \bar z = 8 = 2\operatorname{Re}(z), \quad z - \bar z = -6i = 2i\operatorname{Im}(z), \quad z\bar z = 16 + 9 = 25 = |z|^2.$

(B1 each — three independent results; note $z\bar z$ is real and equals $|z|^2$ .) These three identities are worth internalising: they let you extract the real part, imaginary part and squared modulus of any $z$ purely from $z$ and $\bar z$ , without re-reading off $a$ and $b$ . They are used constantly when proving locus equations and when manipulating expressions that mix $z$ with $\bar z$ .

Example 2 — solving a cubic from one complex root

$p(x) = x^3 - 5x^2 + 11x - 15$ has root $x = 1 + 2i$ . Find all roots.

$\text{Real coefficients} \Rightarrow x = 1 - 2i \text{ is also a root.} \quad (\text{M1 state conjugate})$ $\text{Quadratic factor: } x^2 - (z + \bar z)x + z\bar z = x^2 - 2x + 5. \quad (\text{M1 form factor, A1})$ $p(x) = (x^2 - 2x + 5)(x - c); \text{ constant term } 5(-c) = -15 \Rightarrow c = 3. \quad (\text{M1 divide/compare, A1})$

Roots: $1 + 2i,\ 1 - 2i,\ 3$ . Check: $(x^2 - 2x + 5)(x - 3)$ expands to $x^3 - 5x^2 + 11x - 15$ ✓.

Example 3 — a quartic from one purely imaginary root

$q(x) = x^4 - 2x^3 + 6x^2 - 8x + 8$ has root $x = 2i$ . Find all roots.

$\Rightarrow x = -2i \text{ is a root; factor } x^2 - (2i - 2i)x + (2i)(-2i) = x^2 + 4. \quad (\text{M1 M1})$ $q(x) = (x^2 + 4)(x^2 - 2x + 2) \quad (\text{M1 division, A1})$ $x^2 - 2x + 2 = 0 \Rightarrow \Delta = 4 - 8 = -4 \Rightarrow x = \frac{2 \pm 2i}{2} = 1 \pm i. \quad (\text{A1})$

All four roots: $2i,\ -2i,\ 1 + i,\ 1 - i$ . (Check: the two real quadratic factors $x^2+4$ and $x^2-2x+2$ multiply back to $q(x)$ ✓.)

Example 4 — finding the quotient by comparing coefficients

Solve $p(x) = 2x^3 - 9x^2 + 14x - 5 = 0$ given that $2 - i$ is a root.

The conjugate $2 + i$ is also a root, giving the real quadratic factor

$x^2 - (z + \bar z)x + z\bar z = x^2 - 4x + 5. \quad (\text{M1})$

Since the leading coefficient is $2$ , write $p(x) = (x^2 - 4x + 5)(2x + k)$ and find $k$ by comparing coefficients:

$(x^2 - 4x + 5)(2x + k) = 2x^3 + (k - 8)x^2 + (20 - 4k)x + 5k. \quad (\text{M1 expand})$

Comparing the constant term $5k = -5$ gives $k = -1$ (and the $x^2$ coefficient checks: $k - 8 = -9$ ✓). So the third factor is $2x - 1$ , giving the real root $x = \tfrac12$ . (A1.)

The three roots are $2 - i,\ 2 + i,\ \tfrac12$ . (Comparing coefficients is often quicker and cleaner than long division when the leading coefficient is not $1$ — choose whichever you find more reliable.)

Example 5 — a quartic with two conjugate pairs

Solve $x^4 + 13x^2 + 36 = 0$ .

This is a quadratic in $x^2$ : let $u = x^2$ , so $u^2 + 13u + 36 = 0$ , which factors as $(u + 4)(u + 9) = 0$ . (M1 spot the quadratic-in- $x^2$ ; A1 factor.) Hence $x^2 = -4$ or $x^2 = -9$ , giving

$x = \pm 2i \quad \text{or} \quad x = \pm 3i. \quad (\text{A1 all four roots})$

All four roots are purely imaginary, forming two conjugate pairs $\{2i, -2i\}$ and $\{3i, -3i\}$ — fully consistent with a real-coefficient quartic having an even number of non-real roots.

Specimen-style exam question

(specimen-style — not from any past paper)

(a) Show that $x = -1 + 3i$ is a root of $f(x) = x^3 + 6x - 20$ … in fact, form the cubic with real coefficients whose roots are $2$ and $-1 + 3i$ .

The third root is the conjugate $-1 - 3i$ . The conjugate-pair factor is

$(x - (-1+3i))(x - (-1-3i)) = (x + 1)^2 + 9 = x^2 + 2x + 10.$

Multiplying by the real-root factor $(x - 2)$ :

$f(x) = (x - 2)(x^2 + 2x + 10) = x^3 + 2x^2 + 10x - 2x^2 - 4x - 20 = x^3 + 6x - 20.$

(Verify: $f(2) = 8 + 12 - 20 = 0$ ✓, confirming $x = 2$ is a root.)

Complex Conjugates and Polynomial Roots

Complex Conjugates and Polynomial Roots

Where this sits in AQA 7367

Core theory: the conjugate and its algebra

The conjugate root theorem

The real quadratic factor

The standard solving strategy

Worked examples (with mark scheme)

Example 1 — conjugate identities

Example 2 — solving a cubic from one complex root

Example 3 — a quartic from one purely imaginary root

Example 4 — finding the quotient by comparing coefficients

Example 5 — a quartic with two conjugate pairs

Specimen-style exam question

Synoptic links

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