Complex numbers (WACE Year 12 Mathematics Specialist Unit 3)

What this dot point is asking

SCSA wants you to move fluently between Cartesian form $z = x + iy$ and polar (modulus-argument) form, plot numbers on the Argand plane, carry out all four arithmetic operations, use the conjugate, and apply de Moivre's theorem to find powers and the $n$ roots of a complex number.

Cartesian form and arithmetic

Write $z = x + iy$ where $x = \operatorname{Re}(z)$ and $y = \operatorname{Im}(z)$, and $i^2 = -1$. Addition and subtraction act component-wise. Multiplication uses the distributive law with $i^2 = -1$:

The conjugate of $z = x + iy$ is $\bar z = x - iy$. The product $z\bar z = x^2 + y^2 = |z|^2$ is real and non-negative, which is the key to division: multiply numerator and denominator by the conjugate of the denominator.

The Argand plane, modulus and argument

Plot $z = x + iy$ as the point $(x, y)$. The modulus $|z| = \sqrt{x^2 + y^2}$ is the distance from the origin. The argument $\arg z$ is the angle measured anticlockwise from the positive real axis. The principal argument $\operatorname{Arg} z$ lies in $(-\pi, \pi]$. Always check the quadrant: a bare $\tan^{-1}(y/x)$ can land in the wrong half-plane.

Polar form and de Moivre's theorem

In polar form $z = r\,\text{cis}\,\theta$ where $\text{cis}\,\theta = \cos\theta + i\sin\theta$. Multiplication multiplies moduli and adds arguments; division divides moduli and subtracts arguments:

For the $n$ distinct $n$th roots of $w = R\,\text{cis}\,\phi$, solve $z^n = w$:

The roots all have modulus $R^{1/n}$ and are spaced $\tfrac{2\pi}{n}$ apart, so they sit on the vertices of a regular polygon.

Converting between forms

Moving between Cartesian and polar form is a core SCSA skill. To go from Cartesian $z = x + iy$ to polar, compute the modulus $r = \sqrt{x^2 + y^2}$ and the argument $\theta$ by considering the quadrant of $(x, y)$. To go from polar back to Cartesian, evaluate $x = r\cos\theta$ and $y = r\sin\theta$. For example $z = 4\,\text{cis}\,\tfrac{\pi}{6}$ becomes $x = 4\cos\tfrac{\pi}{6} = 4 \times \tfrac{\sqrt{3}}{2} = 2\sqrt{3}$ and $y = 4\sin\tfrac{\pi}{6} = 4 \times \tfrac{1}{2} = 2$, so $z = 2\sqrt{3} + 2i$. Knowing the exact values of $\cos$ and $\sin$ at the standard angles $\tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}$ is essential for the calculator-free section.

When polar form is the better tool

Polar form turns multiplication, division, powers and roots into simple operations on the modulus and argument, so it is the natural choice whenever a question raises a complex number to a power or asks for roots. Cartesian form is better for addition and subtraction, which act componentwise but are awkward in polar form. A common exam strategy is to add or subtract in Cartesian form, then convert to polar form before applying de Moivre's theorem.