What is wrong centre sign?

|z - a| = r is centred at +a; for |z + 3 - 2i| rewrite as |z - (-3 + 2i)| so the centre is (-3, 2).

Describe (z - i) = π2. [2 marks]

SingaporeFurther MathsSyllabus dot point

How do equations and inequalities involving the modulus and argument define loci and regions on the Argand diagram?

Sketch loci and regions in the Argand diagram defined by conditions on the modulus and argument of a complex number

A focused answer to the H2 Further Mathematics outcome on loci in the complex plane. Circles from modulus conditions, perpendicular bisectors from equal-distance conditions, half-lines from argument conditions, and shading regions defined by inequalities.

Generated by Claude Opus 4.811 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 interpret conditions on the modulus and argument of a complex number as geometric loci and regions in the Argand diagram, to sketch them accurately, and to read off quantities such as the greatest or least modulus. The three core loci are the circle, the perpendicular bisector, and the half-line, and inequalities of the same forms give regions to shade.

The answer

Distance interpretation of the modulus

The key idea is that $|z - a|$ is the distance from the point $z$ to the fixed point $a$ on the Argand diagram. Every locus in this topic is built from this reading.

Circles: a fixed distance from a point

|z - a| = r \quad (r > 0)

is the set of points at distance $r$ from $a$ , a circle of radius $r$ centred at the point $a$ . The inequality $|z - a| \leq r$ shades the closed disc, and $|z - a| > r$ the exterior.

Perpendicular bisectors: equal distances from two points

|z - a| = |z - b|

is the set of points equidistant from $a$ and $b$ , the perpendicular bisector of the segment joining them. The inequality $|z - a| < |z - b|$ shades the half-plane nearer to $a$ .

Half-lines: a fixed argument

\arg(z - a) = \theta

is a half-line (ray) starting from the point $a$ (excluded) and making angle $\theta$ with the positive real direction. It is a ray, not a full line, because the argument fixes the direction. A range $\alpha \leq \arg(z - a) \leq \beta$ shades the sector between two rays from $a$ .

Reading off greatest and least values

For a circle locus, the greatest and least distances from a fixed external point $P$ are (distance $P$ to centre) $\pm$ radius. The same "centre plus or minus radius" reasoning gives the greatest or least modulus, or the extreme value of $\arg z$ , after a quick sketch.

Worked example

A complex number $z$ satisfies both $|z - 2i| \leq 3$ and $0 \leq \arg z \leq \dfrac{\pi}{4}$ . Sketch the region and find the point of greatest modulus in it.

Step 1: Interpret the first condition

$|z - 2i| \leq 3$ is the closed disc of radius $3$ centred at $(0, 2)$ (the point $2i$ ).

Step 2: Interpret the second condition

$0 \leq \arg z \leq \dfrac{\pi}{4}$ is the sector from the positive real axis up to the ray at $45^\circ$ , both rays starting at the origin.

Step 3: Combine

The region is the part of the disc lying inside that sector: a lens-shaped piece bounded by the two rays and the arc of the circle.

Step 4: Locate the greatest modulus

The greatest $|z|$ in the disc alone is at distance (centre distance from origin) + radius $= 2 + 3 = 5$ , in the direction from the origin through the centre, that is straight up the imaginary axis. But that direction ( $\arg z = \tfrac{\pi}{2}$ ) is outside the sector.

Step 5: Take the boundary constraint into account

Constrained to $\arg z \leq \dfrac{\pi}{4}$ , the farthest point is where the ray $\arg z = \dfrac{\pi}{4}$ meets the circle on its far side. Substituting $z = t\,\mathrm{e}^{i\pi/4}$ into $|z - 2i| = 3$ and taking the larger root of the resulting quadratic gives that intersection, which is the point of greatest modulus in the region.

Common traps

Drawing a full line for an argument condition: $\arg(z - a) = \theta$ is a half-line (ray) from $a$ , not a line through it; the opposite direction has argument $\theta \pm \pi$ .
Including the vertex of a half-line: The point $a$ itself has no defined argument for $z - a$ , so it is excluded (open) from $\arg(z - a) = \theta$ .
Wrong centre sign: $|z - a| = r$ is centred at $+a$ ; for $|z + 3 - 2i|$ rewrite as $|z - (-3 + 2i)|$ so the centre is $(-3, 2)$ .
Forgetting the boundary in optimisation: The unconstrained extreme of $|z|$ may lie outside an additional region; check whether the extra inequality moves the optimum to a boundary.
Shading the wrong side: For $|z - a| < |z - b|$ , shade the half-plane nearer $a$ ; test a point if unsure.

Examples in context

Example 1. Tolerance regions in engineering. A specification that a complex impedance must lie within a given distance of a target value is exactly $|z - a| \leq r$ , a disc. The Argand diagram turns a tolerance into a region to design within.

Example 2. Equal-signal boundaries. The set of points equidistant from two transmitters is the perpendicular bisector $|z - a| = |z - b|$ ; one side receives a stronger signal from $a$ , the other from $b$ , the geometric basis of coverage boundaries.

Try this

Q1. Describe the locus $|z - 5| = 3$ . [1 mark]

Cue. A circle of radius $3$ centred at the point $(5, 0)$ .

Q2. What locus is given by $|z - 2| = |z - 6|$ ? [1 mark]

Cue. The perpendicular bisector of the points $(2, 0)$ and $(6, 0)$ , namely the line $x = 4$ .

Q3. Describe $\arg(z - i) = \dfrac{\pi}{2}$ . [2 marks]

Cue. A half-line starting at the point $(0, 1)$ (excluded), pointing vertically upward.

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.

Original5 marksSketch on an Argand diagram the locus of points

z

satisfying

|z - 3 - 4i| = 2

, and state the greatest value of

|z|

on this locus.

Show worked answer →

$|z - (3 + 4i)| = 2$ is the set of points at distance $2$ from the fixed point $3 + 4i$ . This is a circle of radius $2$ centred at $(3, 4)$ .

The greatest value of $|z|$ is the distance from the origin to the farthest point of the circle, which is the distance from the origin to the centre plus the radius. Distance to the centre: $\sqrt{3^2 + 4^2} = 5$ . So the greatest $|z| = 5 + 2 = 7$ .

Markers reward identifying the circle of radius $2$ centred at $(3, 4)$ , a correct sketch, and the greatest modulus as (distance to centre) + radius $= 7$ .

Original6 marksDescribe and sketch the locus given by

|z - 1| = |z + i|

, and find its Cartesian equation.

Show worked answer →

$|z - 1| = |z - (-i)|$ is the set of points equidistant from $1$ (the point $(1, 0)$ ) and $-i$ (the point $(0, -1)$ ). This is the perpendicular bisector of the segment joining $(1, 0)$ and $(0, -1)$ .

To find its equation, let $z = x + iy$ . Then $|z - 1|^2 = (x - 1)^2 + y^2$ and $|z + i|^2 = x^2 + (y + 1)^2$ . Setting them equal:

(x - 1)^2 + y^2 = x^2 + (y + 1)^2.

Expand:

x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2y + 1

, so

-2x = 2y

, that is

y = -x

The locus is the line $y = -x$ , the perpendicular bisector of the segment.

Markers reward recognising the perpendicular bisector, squaring both moduli, and simplifying to the line $y = -x$ .