Arithmetic and algebra of complex numbers in Cartesian form $z = a + bi$ and polar form $z = r\,\mathrm{cis}\,\theta$, the modulus and argument, conjugates, and representation of complex numbers and their operations on the Argand plane

What this dot point is asking

VCAA wants you to work confidently with complex numbers in both Cartesian form $z = a + bi$ and polar form $z = r\,\mathrm{cis}\,\theta$, to compute the modulus and argument, to use the conjugate, and to plot complex numbers and visualise their arithmetic on the Argand plane. This material underpins De Moivre's theorem and the factorisation of polynomials over the complex numbers, so fluency here pays off across the whole complex-numbers strand.

Cartesian form and the Argand plane

A complex number is written $z = a + bi$ where $a, b \in \mathbb{R}$ and $i$ is the imaginary unit with $i^2 = -1$. We call $a = \mathrm{Re}(z)$ the real part and $b = \mathrm{Im}(z)$ the imaginary part (note the imaginary part is the real number $b$, not $bi$).

The Argand plane represents $z = a + bi$ as the point $(a, b)$, with the horizontal axis the real axis and the vertical axis the imaginary axis. Addition of complex numbers corresponds to vector addition of the position vectors: $(a_1 + b_1 i) + (a_2 + b_2 i) = (a_1 + a_2) + (b_1 + b_2)i$, which is the parallelogram rule.

Modulus and argument

The modulus $|z|$ is the distance from the origin to the point $(a, b)$:

The argument $\arg(z) = \theta$ is the angle the position vector makes with the positive real axis, measured anticlockwise. It satisfies

Because $\tan$ repeats every $\pi$, you must use the signs of $a$ and $b$ to place $\theta$ in the correct quadrant. The principal argument $\mathrm{Arg}(z)$ is the value in $(-\pi, \pi]$.

Conjugates

The conjugate of $z = a + bi$ is $\bar z = a - bi$, the reflection of $z$ across the real axis. Key properties:

The identity $z\bar z = |z|^2$ is the tool for division: to simplify $\dfrac{z_1}{z_2}$, multiply numerator and denominator by $\bar z_2$ so the denominator becomes the real number $|z_2|^2$.

Polar (cis) form

Using $a = r\cos\theta$ and $b = r\sin\theta$ with $r = |z|$, every nonzero complex number can be written

where $\mathrm{cis}\,\theta$ is shorthand for $\cos\theta + i\sin\theta$. Polar form makes multiplication and division simple:

In words: to multiply, multiply the moduli and add the arguments; to divide, divide the moduli and subtract the arguments. This is why polar form is preferred for powers and roots, while Cartesian form is preferred for addition and subtraction.

Why both forms matter

Cartesian form makes the real and imaginary parts explicit and is the natural setting for adding, subtracting, and reading off coordinates on the Argand plane. Polar form encodes the geometry of multiplication: every multiplication is a rotation (by the argument) combined with a scaling (by the modulus). Multiplying by $i = \mathrm{cis}\,\frac{\pi}{2}$, for example, rotates a point $90^\circ$ anticlockwise without changing its distance from the origin. Choosing the right form for the operation is the single biggest efficiency gain on exam questions.

Examples in context

Example 1. Square of a modulus. For $z = 3 + 4i$, $z\bar z = (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 + 16 = 25 = |z|^2$, confirming $|z| = 5$.

Example 2. Rationalising a quotient. $\dfrac{2 + i}{1 - i} = \dfrac{(2 + i)(1 + i)}{(1 - i)(1 + i)} = \dfrac{2 + 2i + i + i^2}{1 + 1} = \dfrac{1 + 3i}{2} = \tfrac{1}{2} + \tfrac{3}{2}i$.

Try this

Q1. Write $z = -1 + \sqrt{3}\,i$ in polar form. [2 marks]

• Cue. $|z| = \sqrt{1 + 3} = 2$; second quadrant so $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$; $z = 2\,\mathrm{cis}\,\frac{2\pi}{3}$.

Q2. Simplify $\dfrac{4}{3 + i}$ to Cartesian form. [2 marks]

• Cue. Multiply by $\frac{3 - i}{3 - i}$: $\frac{4(3 - i)}{10} = \frac{12 - 4i}{10} = \frac{6}{5} - \frac{2}{5}i$.

Q3. If $z_1 = 2\,\mathrm{cis}\,\frac{\pi}{3}$ and $z_2 = 3\,\mathrm{cis}\,\frac{\pi}{6}$, find $z_1 z_2$. [2 marks]

• Cue. Multiply moduli, add arguments: $6\,\mathrm{cis}\,\frac{\pi}{2} = 6i$.