What is wrong side for an inequality region?

Test a point to decide which side of a boundary the region lies on.

Describe the locus (z - 1) = π4. [2 marks]

SEAB Original-style practice: Describe and sketch the locus of z satisfying |z - 2| = |z - 6i|.

The condition says z is equidistant from the points 2 (that is 2 + 0i) and 6i (that is 0 + 6i). The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them. The midpoint of (2, 0) and (0, 6) is (1, 3); the segment has gradient 6 - 00 - 2 = -3, so the perpendicular bisector has gradient 13 and passes through (1, 3): y - 3 = 13(x - 1). Markers reward recognising the equidistance condition, the perpendicular-bisector description, and a correct line through the midpoint.

SingaporeMathsSyllabus dot point

How do complex numbers describe geometry in the Argand diagram, and what loci do conditions on modulus and argument define?

Represent complex numbers on an Argand diagram and identify and sketch loci defined by conditions on the modulus and argument

A focused answer to the H2 Mathematics outcome on the Argand diagram and loci. Plotting complex numbers, the geometric meaning of modulus and argument, and sketching circles, perpendicular bisectors, half-lines and regions.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to plot complex numbers on an Argand diagram, interpret the modulus and argument geometrically, and identify and sketch the loci defined by conditions such as $|z - a| = r$ , $|z - a| = |z - b|$ and $\arg(z - a) = \alpha$ , including regions defined by inequalities.

The answer

The Argand diagram

A complex number $z = x + iy$ is plotted as the point $(x, y)$ . The horizontal axis is the real axis and the vertical axis the imaginary axis. Adding complex numbers corresponds to vector addition of their position vectors.

Modulus and argument as distance and angle

$|z - a|$ is the distance between the points $z$ and $a$ in the plane.
$\arg(z - a)$ is the angle that the vector from $a$ to $z$ makes with the positive real direction.

These two readings unlock every locus.

The standard loci

Circle: $|z - a| = r$ is the circle centre $a$ , radius $r$ (constant distance from a fixed point).
Perpendicular bisector: $|z - a| = |z - b|$ is the set equidistant from $a$ and $b$ , the perpendicular bisector of the segment $ab$ .
Half-line: $\arg(z - a) = \alpha$ is a ray (half-line) starting at $a$ (excluded) at angle $\alpha$ .

Regions from inequalities

Replacing $=$ with $<$ or $>$ gives a region: $|z - a| < r$ is the inside of the circle, $|z - a| > |z - b|$ is the half-plane nearer to $b$ , and a range of arguments gives a wedge. Shade and use dashed or solid boundaries to mark strict or non-strict conditions.

Worked example

Sketch the locus and find the points where it meets the real axis for $|z - 1 - i| = \sqrt{2}$ .

Step 1: Identify the locus

$|z - (1 + i)| = \sqrt{2}$ is the circle with centre $(1, 1)$ and radius $\sqrt{2}$ .

Step 2: Set up the Cartesian equation

With $z = x + iy$ , $(x - 1)^2 + (y - 1)^2 = 2$ .

Step 3: Intersect with the real axis (y = 0)

(x - 1)^2 + (0 - 1)^2 = 2 \implies (x - 1)^2 = 1 \implies x - 1 = \pm 1

Step 4: Solve

$x = 2$ or $x = 0$ . The circle meets the real axis at $(2, 0)$ and $(0, 0)$ .

Step 5: Describe the sketch

A circle centred at $(1, 1)$ of radius $\sqrt{2} \approx 1.41$ , passing through the origin and $(2, 0)$ on the real axis.

Common traps

Reading the centre with the wrong sign: $|z - (3 + 4i)|$ has centre $(3, 4)$ ; the centre is the value subtracted, not its negative.
Confusing a circle with a perpendicular bisector: $|z - a| = r$ (one fixed point, constant distance) is a circle; $|z - a| = |z - b|$ (two fixed points) is a line.
Including the endpoint of a half-line: $\arg(z - a) = \alpha$ starts at $a$ but excludes it, since $\arg(0)$ is undefined; mark it with an open circle.
Forgetting the argument range: A half-line is only one ray at angle $\alpha$ , not the full line; the opposite ray has argument $\alpha + \pi$ .
Wrong side for an inequality region: Test a point to decide which side of a boundary the region lies on.

Examples in context

Example 1. Signal within tolerance. A communications signal whose complex amplitude must stay within a tolerance of a target value $a$ satisfies $|z - a| \leq \epsilon$ , the closed disc centre $a$ radius $\epsilon$ , the geometric picture of an acceptable signal region.

Example 2. Equal-distance boundary. A boundary equidistant from two transmitters at $a$ and $b$ is the perpendicular bisector $|z - a| = |z - b|$ , exactly the locus that separates which transmitter is nearer, a direct mapping of the algebra to a coverage map.

Try this

Q1. Describe the locus $|z| = 3$ . [1 mark]

Cue. A circle centred at the origin with radius $3$ .

Q2. Describe the locus $|z - 2i| = |z + 4|$ . [2 marks]

Cue. Perpendicular bisector of the segment joining $(0, 2)$ and $(-4, 0)$ .

Q3. Describe the locus $\arg(z - 1) = \dfrac{\pi}{4}$ . [2 marks]

Cue. A half-line starting at $(1, 0)$ (excluded), at $45^\circ$ above the positive real direction.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksSketch on an Argand diagram the locus of points

z

satisfying

|z - 3 - 4i| = 2

, and state its geometric description.

Show worked answer →

$|z - (3 + 4i)| = 2$ means the distance from $z$ to the fixed point $3 + 4i$ is constant and equal to $2$ .

This is a circle with centre $(3, 4)$ and radius $2$ .

The sketch shows a circle centred at $(3, 4)$ passing through points such as $(5, 4)$ , $(1, 4)$ , $(3, 6)$ and $(3, 2)$ .

Markers reward interpreting the modulus as a distance, identifying the centre and radius, and a correct circular locus.

Original5 marksDescribe and sketch the locus of

z

satisfying

|z - 2| = |z - 6i|

Show worked answer →

The condition says $z$ is equidistant from the points $2$ (that is $2 + 0i$ ) and $6i$ (that is $0 + 6i$ ).

The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.

The midpoint of $(2, 0)$ and $(0, 6)$ is $(1, 3)$ ; the segment has gradient $\dfrac{6 - 0}{0 - 2} = -3$ , so the perpendicular bisector has gradient $\dfrac{1}{3}$ and passes through $(1, 3)$ : $y - 3 = \tfrac{1}{3}(x - 1)$ .

Markers reward recognising the equidistance condition, the perpendicular-bisector description, and a correct line through the midpoint.