The arithmetic of complex numbers, the complex conjugate and division, the Argand diagram, and solving quadratic, cubic and quartic equations with complex roots that occur in conjugate pairs.

What this dot point is asking

OCR wants you to do arithmetic with complex numbers fluently: add, subtract and multiply them like binomials with $i^2 = -1$, divide by multiplying numerator and denominator by the conjugate of the denominator, plot complex numbers on an Argand diagram, and use the fact that the complex roots of a real polynomial occur in conjugate pairs to solve quadratic, cubic and quartic equations.

Arithmetic and the imaginary unit

The imaginary unit satisfies $i^2 = -1$. A complex number $z = x + yi$ combines a real part $\operatorname{Re}(z) = x$ and an imaginary part $\operatorname{Im}(z) = y$. Addition and subtraction act on the parts separately, and multiplication follows the distributive law with $i^2 = -1$.

The conjugate and division

The complex conjugate $z^* = x - yi$ reflects $z$ in the real axis. Its key property is that $zz^* = x^2 + y^2 = |z|^2$ is always real and non-negative, which is what makes division possible: to divide, multiply numerator and denominator by the conjugate of the denominator, turning the denominator real.

The Argand diagram

The Argand diagram plots $z = x + yi$ as the point $(x, y)$, with the real axis horizontal and the imaginary axis vertical. Addition of complex numbers is then vector addition (place the two displacement arrows nose to tail), subtraction is the displacement between the two points, and the conjugate is the reflection in the real axis. Multiplying by a real number scales the point along its own direction, while multiplying by $i$ rotates it a quarter turn anticlockwise. Sketching $z$ is the first step in any modulus-argument or loci question because it fixes the geometry, and in particular it tells you which quadrant the point lies in, which is exactly what you need to read off the correct argument.

Roots of polynomials in conjugate pairs

For a polynomial with real coefficients, if $z = a + bi$ is a root then so is its conjugate $z^* = a - bi$. The two together give a real quadratic factor, which is the key to solving cubics and quartics once one complex root is known.

So a real cubic with one complex root must have its conjugate as a second root and a single real root, while a real quartic with two complex roots has them as two conjugate pairs.

This arithmetic underpins the modulus-argument form, de Moivre's theorem and loci, all of which build on plotting $z$ and manipulating conjugates.

Try this

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

• Cue. $(2 - 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i$.

Q2. A real quadratic has $3 - 2i$ as a root. Write the quadratic with real coefficients. [3 marks]

• Cue. The other root is $3 + 2i$; sum $= 6$, product $= 9 + 4 = 13$, so $z^2 - 6z + 13 = 0$.