Solve quadratic equations with complex coefficients and factorise polynomials over the complex field, using the conjugate root theorem for real polynomials

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 $\mathbb{C}$.

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 $2uv = q$. There are two square roots, differing only in sign, and the $\pm$ in the formula accounts for both.

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 $P(\bar{\alpha}) = 0$.

A consequence 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.

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.

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.