Subsets of the complex plane: VCE Specialist Mathematics Unit 3

Syllabus dot point

“Description and sketching of subsets of the complex plane defined by conditions on modulus and argument, including circles $|z - z_0| = r$ , perpendicular bisectors $|z - a| = |z - b|$ , rays $\arg(z - z_0) = \alpha$ , and the regions defined by the corresponding inequalities”

6 min readUpdated 7 Oct 2026

Quick answer

Reading $z = x + iy$ as the point $(x, y)$ turns conditions into loci. $|z - z_0| = r$ is a circle of radius $r$ centred at $z_0$ , and $|z - z_0| \le r$ is the closed disc. $|z - a| = |z - b|$ is the perpendicular bisector of the segment joining $a$ and $b$ , since it is the set of points equidistant from the two. $\arg(z - z_0) = \alpha$ is a ray (half line) from $z_0$ at angle $\alpha$ , with the endpoint $z_0$ excluded because its argument is undefined. Inequalities replace the boundary curve with the region on one side of it.

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What this dot point is asking
Circles and discs
Perpendicular bisectors
Rays and arguments
Combining conditions
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to take a condition on a complex number $z$ , expressed through its modulus or argument, and sketch the set of points it describes on the Argand plane. The standard loci are circles, perpendicular bisectors and rays; inequalities turn these boundaries into regions. The skill is translating algebra into geometry and back.

Circles and discs

The modulus $|z - z_0|$ is the distance from $z$ to the fixed point $z_0$ . Setting that distance equal to a constant gives a circle:

|z - z_0| = r \quad\text{is the circle of radius } r \text{ centred at } z_0.

The inequalities describe regions: $|z - z_0| < r$ is the open interior (disc without boundary), $|z - z_0| \le r$ is the closed disc, and $|z - z_0| > r$ is the exterior. To verify algebraically, write $z = x + iy$ and $z_0 = a + ib$ ; then $|z - z_0|^2 = (x - a)^2 + (y - b)^2 = r^2$ , the Cartesian circle.

Perpendicular bisectors

The condition $|z - a| = |z - b|$ says $z$ is equidistant from the fixed points $a$ and $b$ . Geometrically this is the perpendicular bisector of the line segment joining $a$ and $b$ . The corresponding inequality $|z - a| < |z - b|$ is the half plane of points closer to $a$ than to $b$ .

Rays and arguments

The argument $\arg(z - z_0)$ is the direction of the displacement from $z_0$ to $z$ . Fixing it,

\arg(z - z_0) = \alpha

gives a ray: the half line starting at $z_0$ and pointing in the direction $\alpha$ . The point $z_0$ itself is excluded, since the argument of $0$ is undefined; draw an open circle there. A pair of inequalities such as $\frac{\pi}{6} \le \arg(z - z_0) \le \frac{\pi}{3}$ describes the angular sector (wedge) between two rays.

Combining conditions

Many questions intersect two or more loci, for example a disc and a sector, and ask for the overlap region. Sketch each locus, shade the region each inequality allows, and take the common part. Mark boundary curves as solid when the inequality is non-strict ( $\le$ , $\ge$ ) and dashed when strict ( $<$ , $>$ ).

Worked example

Sketching an intersection

Sketch the set of points satisfying both $|z - (1 + i)| \le 2$ and $0 \le \arg(z - (1 + i)) \le \frac{\pi}{2}$ .

First condition: $|z - (1 + i)| \le 2$ is the closed disc of radius $2$ centred at the point $(1, 1)$ . The boundary circle is drawn solid because the inequality is $\le$ .
Second condition: $\arg(z - (1 + i))$ is the direction from the centre $(1, 1)$ to $z$ . Requiring it between $0$ and $\frac{\pi}{2}$ selects the directions from due east round to due north, that is, the upper-right quarter measured from $(1, 1)$ . This is the quarter-disc sector.
Intersection: The region is the quarter of the disc lying up and to the right of the centre: bounded below by the horizontal ray $\arg = 0$ (pointing right from $(1,1)$ ), on the left by the vertical ray $\arg = \frac{\pi}{2}$ (pointing up), and on the outside by the arc of the circle of radius $2$ .

Check a sample point. The point $z = 2 + 2i$ has $z - (1 + i) = 1 + i$ , modulus $\sqrt{2} \le 2$ and argument $\frac{\pi}{4} \in \left[0, \frac{\pi}{2}\right]$ , so it lies in the region, confirming the sketch.

Common exam traps

Including the ray endpoint: On $\arg(z - z_0) = \alpha$ the point $z_0$ is excluded. Mark it with an open circle.
Drawing a full line instead of a ray: An argument condition gives a half line, not the whole line; the opposite direction has argument $\alpha + \pi$ .
Wrong centre sign: $|z - z_0| = r$ is centred at $z_0$ , so $|z + 2 - i| = 3$ is centred at $-2 + i$ , not $2 - i$ .
Solid versus dashed boundaries: Strict inequalities exclude the boundary (dashed); non-strict include it (solid). Examiners check this.

Examples in context

Example 1. $|z| = 3$ is the circle of radius $3$ centred at the origin.

Example 2. $|z - 2| = |z + 2i|$ is the perpendicular bisector of the segment from $2$ to $-2i$ , that is, the points equidistant from $(2, 0)$ and $(0, -2)$ .

Try this

Q1. Describe the locus $|z - 4i| = 5$ . [2 marks]

Cue. Circle of radius $5$ centred at $(0, 4)$ .

Q2. Sketch $\arg(z) = \frac{3\pi}{4}$ . [2 marks]

Cue. Ray from the origin (excluded) into the second quadrant at $135^\circ$ .

Q3. Describe the region $1 \le |z| \le 2$ . [2 marks]

Cue. The closed annulus between circles of radius $1$ and $2$ about the origin.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2025 VCAA-style1 mark

Sketch { z : z times conjugate(z) = 4, z in C } on the Argand plane.

Show worked answer →

Use the identity z times conjugate(z) = |z|^2 for any complex number z.

So the condition z times conjugate(z) = 4 becomes |z|^2 = 4, that is |z| = 2.

This is the set of all complex numbers whose distance from the origin is 2. The graph is a circle centred at the origin O with radius 2.

For the sketch, draw a full circle of radius 2 about the origin, passing through the points 2, -2, 2i and -2i on the axes.

Source: VCAA 2025 Specialist Mathematics Examination 2, Question 2a.

2023 VCAA-style1 mark

Find the equation of the ray that originates at the real root of z^7 - 1 = 0 and passes through the point represented by cis(2pi/7), in the form Arg(z - z0) = theta, where z0 is in C and theta is measured in radians in terms of pi.

Show worked answer →

The real root of z^7 - 1 = 0 is z = 1, so the ray originates there: z0 = 1.

The direction of the ray is the argument of the vector from z0 = 1 to the point cis(2pi/7) = (cos(2pi/7), sin(2pi/7)).

That vector is (cos(2pi/7) - 1) + i sin(2pi/7). Using cos(t) - 1 = -2sin^2(t/2) and sin(t) = 2sin(t/2)cos(t/2) with t = 2pi/7, the vector is proportional to (-sin(pi/7), cos(pi/7)), whose argument is pi/2 + pi/7.

Hence theta = pi/2 + pi/7 = 9pi/14, and the ray is Arg(z - 1) = 9pi/14.

Geometrically this is the standard result that a chord of the unit circle from angle 0 to angle alpha points at angle pi/2 + alpha/2.

Source: VCAA 2023 Specialist Mathematics Examination 2, Question 2 part d.ii.