Perform arithmetic with complex numbers in Cartesian form, use the conjugate, and solve polynomial equations including the use of conjugate root pairs

What this dot point is asking

SEAB wants you to do arithmetic with complex numbers in Cartesian form $a + bi$, use the complex conjugate (especially to divide), and solve polynomial equations with complex roots, exploiting the fact that complex roots of a real polynomial come in conjugate pairs.

The answer

The imaginary unit and Cartesian form

The imaginary unit satisfies $i^2 = -1$. A complex number is written $z = a + bi$ with real part $a = \operatorname{Re}(z)$ and imaginary part $b = \operatorname{Im}(z)$.

Arithmetic

• Addition and subtraction: combine real and imaginary parts separately.
• Multiplication: expand as binomials and replace $i^2$ by $-1$.
• Division: multiply numerator and denominator by the conjugate of the denominator to make the denominator real.

The conjugate

The conjugate of $z = a + bi$ is $z^* = a - bi$ (reflection in the real axis). Key facts: $z z^* = a^2 + b^2 = |z|^2$ is real and non-negative, and $z + z^* = 2a$ is real. The conjugate is the tool that rationalises a complex denominator.

Solving polynomial equations

A real quadratic with negative discriminant has a conjugate pair of complex roots, found from the quadratic formula with $\sqrt{-k} = i\sqrt{k}$. For higher-degree real polynomials, the conjugate root theorem says non-real roots occur in conjugate pairs, so knowing one gives a second for free, and the sum and product of a pair give a real quadratic factor.

Reconstructing the real quadratic factor from a complex root

When a real polynomial has a known complex root, the fastest way to extract a real factor is to use the sum and product of the conjugate pair, rather than expanding two complex linear factors. If $a + bi$ is a root, the pair has sum $2a$ and product $a^2 + b^2$, so the real quadratic factor is $z^2 - (2a)z + (a^2 + b^2)$. For the root $2 - 3i$, the factor is $z^2 - 4z + 13$. This shortcut turns "factor a quartic given one complex root" into a quick subtraction problem: divide the polynomial by this real quadratic to reveal the remaining factor, all without ever multiplying complex numbers together.

Equating real and imaginary parts to solve for unknowns

A single complex equation carries two real equations, because two complex numbers are equal only when both their real and their imaginary parts match. So an equation such as $(x + yi)(2 - i) = 5 + 5i$ becomes two simultaneous real equations once you expand and separate parts. Expanding gives $(2x + y) + (-x + 2y)i = 5 + 5i$, so $2x + y = 5$ and $-x + 2y = 5$, solved to $x = 1$, $y = 3$. Splitting one complex equation into its real and imaginary components is a routine H2 technique for finding unknown real numbers hidden inside a complex relationship.

Examples in context

Example 1. AC circuit impedance. Electrical engineers represent impedance as a complex number $Z = R + iX$; dividing the complex voltage by $Z$ to find current uses exactly the conjugate-multiplication division technique, the everyday workhorse of circuit analysis.

Example 2. Factoring a quartic. A real quartic with roots $1 \pm 2i$ and $3 \pm i$ factors into two real quadratics $(z^2 - 2z + 5)(z^2 - 6z + 10)$, each built from a conjugate pair's sum and product, showing how conjugate pairs keep the coefficients real.

Try this

Q1. Express $(2 + i)(3 - 2i)$ in the form $a + bi$. [2 marks]

• Cue. $6 - 4i + 3i - 2i^2 = 6 - i + 2 = 8 - i$.

Q2. Find the conjugate of $-5 + 4i$ and the value of $z z^*$. [2 marks]

• Cue. Conjugate $-5 - 4i$; $z z^* = 25 + 16 = 41$.

Q3. Given $1 + i$ is a root of a real quadratic $z^2 + bz + c = 0$, find $b$ and $c$. [3 marks]

• Cue. Other root $1 - i$; sum $2 = -b$ so $b = -2$; product $2 = c$.