Polynomials over the complex field in HSC Mathematics Extension 2: solving quadratics with complex coefficients, the conjugate root theorem, and complete factorisation

What this dot point is asking

NESA wants you to work with polynomials over the complex numbers. You must solve quadratic equations whose coefficients may themselves be complex, apply the conjugate root theorem to real polynomials (complex roots occur in conjugate pairs), use the fundamental theorem of algebra to count roots, and factorise a polynomial completely into linear factors over $\mathbb{C}$. The conjugate root theorem is the workhorse: it is what lets a single supplied complex root unlock a whole real polynomial.

Quadratics with complex coefficients

The quadratic formula

applies to any quadratic $az^2 + bz + c = 0$, even when $a, b, c$ are complex. The new feature is that $\sqrt{b^2 - 4ac}$ is a complex square root. To find $\sqrt{w}$ for a complex $w = p + qi$, set $\sqrt{w} = u + vi$ and solve $u^2 - v^2 = p$ and $2uv = q$ (often together with the modulus equation $u^2 + v^2 = |w|$, which makes the algebra fall out quickly). There are two square roots, differing only in sign, and the $\pm$ in the formula accounts for both.

When the coefficients are real but the discriminant is negative, the formula still works directly: $\sqrt{b^2 - 4ac} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}$, and the two roots come out as a conjugate pair immediately.

The conjugate root theorem

When a polynomial $P(z)$ has only real coefficients, its non-real roots come in conjugate pairs. That is, if $\alpha = a + bi$ (with $b \neq 0$) satisfies $P(\alpha) = 0$, then $P(\bar{\alpha}) = 0$ also. The reason is that conjugation respects addition and multiplication, so $\overline{P(\alpha)} = P(\bar{\alpha})$; since $P(\alpha) = 0$ and $\bar{0} = 0$, we get $P(\bar{\alpha}) = 0$.

Geometrically, the roots of a real polynomial are symmetric about the real axis: every non-real root is mirrored by its conjugate. The figure builds this picture for the cubic $z^3 - 5z^2 + 17z - 13$, whose roots are $1$, $2 + 3i$ and $2 - 3i$.

Stage 1, a single non-real root. Suppose a question tells you that $2 + 3i$ is a root of a polynomial with real coefficients. Plot it in the upper half-plane.

Stage 2, the conjugate must appear. Because the coefficients are real, the conjugate $2 - 3i$ is forced to be a root too. It is the mirror image of $2 + 3i$ in the real axis, so the two roots sit symmetrically above and below.

Stage 3, the remaining root is real. The pair $2 \pm 3i$ contributes a real quadratic factor $z^2 - 4z + 13$. A cubic has three roots, so the third must be real (a lone non-real root could not pair up). Here it is $z = 1$, sitting on the real axis.

A consequence of the pairing is that each conjugate pair $\alpha, \bar{\alpha}$ contributes a real quadratic factor

which has real coefficients. This explains why every real polynomial factorises over the reals into linear and irreducible quadratic factors, and why a real polynomial of odd degree must have at least one real root.

The fundamental theorem of algebra

The fundamental theorem of algebra states that every polynomial of degree $n \ge 1$ with complex coefficients has at least one complex root. By repeatedly factoring out roots, a degree $n$ polynomial has exactly $n$ roots in $\mathbb{C}$ counted with multiplicity, and so factorises completely as

where $a$ is the leading coefficient and $z_1, \ldots, z_n$ are the roots. Over $\mathbb{C}$ there are no irreducible quadratics: everything splits into linear factors.

Sum and product of roots (Vieta)

Vieta's formulas relate the roots to the coefficients and are often the fastest route once a pair of roots is known. For a monic polynomial $z^n + c_{n-1}z^{n-1} + \dots + c_0$, the sum of the roots is $-c_{n-1}$ and the product of the roots is $(-1)^n c_0$. As the 2023 quartic question shows, knowing one conjugate pair plus these two relations can pin down the remaining roots without any division.

Strategy for complete factorisation

Given a real polynomial, find one root (by inspection, the rational root test, or a supplied value). If it is non-real, immediately pair it with its conjugate to obtain a real quadratic factor, then divide. Continue until only linear factors and at most one quadratic remain, then solve the remaining quadratic by formula. If the question asks for factorisation over $\mathbb{C}$, finish by splitting every real quadratic into its two linear conjugate factors.

How exam questions ask about complex polynomials

• "Solve $az^2 + bz + c = 0$ ... in Cartesian form": apply the quadratic formula; if the discriminant is negative (real coefficients) the roots are a conjugate pair $p \pm qi$.
• "Find the square roots of $w$": set $(u + iv)^2 = w$, equate real and imaginary parts, and solve (add the modulus equation to speed it up).
• "$\alpha$ is a zero of $P(z)$ with real coefficients. Explain why $\bar\alpha$ is also a zero": cite the conjugate root theorem (conjugation preserves $+$ and $\times$, so $\overline{P(\alpha)} = P(\bar\alpha) = 0$).
• "Find the remaining zeros / factorise $P(z)$": pair the given root with its conjugate to get a real quadratic, then divide or use Vieta's sum and product.
• "Factorise $P(z)$ over $\mathbb{C}$ / over $\mathbb{R}$": over $\mathbb{C}$ split into all linear factors; over $\mathbb{R}$ leave conjugate pairs combined as real quadratics.
• "Hence solve $\ldots$": a square-root or factor result from an earlier part is meant to be reused, often via completing the square.