Represent complex numbers in Cartesian, polar (modulus-argument) and exponential form, convert between forms, and apply de Moivre's theorem to compute powers and roots, locating results on the complex plane

What this dot point is asking

QCAA wants you to move fluently between the Cartesian form $z = x + iy$, the polar form $z = r(\cos\theta + i\sin\theta)$ and the exponential form $z = re^{i\theta}$, and to use de Moivre's theorem to raise complex numbers to powers and to extract roots. This material is central to IA2 and the external assessment, where geometric interpretation on the Argand plane earns marks alongside algebra.

The answer

Modulus and argument

For $z = x + iy$, the modulus is the distance from the origin,

and the argument $\arg z = \theta$ is the angle measured anticlockwise from the positive real axis. The principal argument is taken in $(-\pi, \pi]$. Always check the quadrant: $\tan\theta = y/x$ alone does not fix the quadrant, so use the signs of $x$ and $y$.

Polar and exponential form

Polar (modulus-argument) form is

often abbreviated $r\,\text{cis}\,\theta$. Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ gives the exponential form

These are the same number written three ways. Exponential form makes multiplication trivial because index laws apply.

Multiplication and division as rotations

Multiplying multiplies moduli and adds arguments:

Geometrically, multiplying by $z_2$ scales by $r_2$ and rotates by $\theta_2$. Multiplying by $i = e^{i\pi/2}$ is a $90^\circ$ anticlockwise rotation.

De Moivre's theorem

For integer $n$,

equivalently $(re^{i\theta})^n = r^n e^{in\theta}$. This turns the messy expansion of a power into a single scaling of modulus and multiplication of argument. De Moivre's theorem can be proved for positive integers by mathematical induction, linking back to Topic 1.

Roots of a complex number

To solve $w^n = z$ where $z = re^{i\theta}$, the $n$ distinct roots are

All roots have the same modulus $r^{1/n}$, so they lie on a circle of that radius, equally spaced by $2\pi/n$ radians. Adding $2\pi k$ to the argument before dividing is essential; otherwise you find only one of the $n$ roots. The $n$-th roots of unity are the special case $z = 1$, giving roots at angles $2\pi k/n$.

Why exponential form helps

Because $e^{i\theta}$ obeys index laws, powers, products and roots reduce to arithmetic on the exponent. Converting a hard Cartesian computation such as $(1+i)^{10}$ into polar form, applying de Moivre, then converting back is far faster and less error-prone than binomial expansion.