What is interpret the first condition?

|z - 2| ≤ 2 is the closed disc centred at z_0 = 2 (the point (2, 0)) with radius 2. The boundary circle (x - 2)^2 + y^2 = 4 is solid because the inequality is inclusive.

What is interpret the second condition?

With z = x + iy, Im(z) = y, so y ≥ 1 is the closed half-plane on and above the horizontal line y = 1, drawn solid.

What is find the overlap?

Shade the part of the disc lying on or above y = 1. The line y = 1 cuts the circle where (x - 2)^2 + 1 = 4, so (x - 2)^2 = 3 and x = 2 3.

What is describe the result?

The region is the circular segment of the disc above the chord from (2 - 3,\, 1) to (2 + 3,\, 1), up to the top of the circle at (2, 2). Both boundaries are solid since both inequalities are inclusive.

What is check a point?

Take z = 2 + 1.5i: |z - 2| = 1.5 ≤ 2 holds, and Im(z) = 1.5 ≥ 1 holds, so this interior point is correctly in the shaded region.

QLDSpecialist MathematicsSyllabus dot point

Topic 3: Complex numbers

Identify and sketch subsets of the complex plane determined by relations involving modulus, argument, distance and inequalities, including lines, circles, perpendicular bisectors, rays and regions

A focused answer to the QCE Specialist Mathematics Unit 3 dot point on subsets of the complex plane. Covers circles and discs from modulus relations, perpendicular bisectors from equal-distance relations, rays from argument conditions, and combining inequalities into regions, with a verified worked example and the boundary-inclusion trap.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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What this dot point is asking

QCAA wants you to translate a condition on a complex number $z$ into a curve or region on the Argand plane and sketch it accurately. The conditions use modulus (distance), argument (angle) and inequalities. Recognising that $|z - z_0|$ is a distance and $\arg(z - z_0)$ is a direction lets you convert algebra into geometry quickly.

The answer

Modulus as distance

For $z = x + iy$ , the expression $|z - z_0|$ is the distance from the point $z$ to the fixed point $z_0$ on the plane. So

|z - z_0| = r

is a circle of radius $r$ centred at $z_0$ . The inequality $|z - z_0| \leq r$ is the closed disc (interior plus boundary), and $|z - z_0| > r$ is the exterior.

Perpendicular bisector from equal distances

The condition $|z - z_1| = |z - z_2|$ says $z$ is equidistant from $z_1$ and $z_2$ . The set of such points is the perpendicular bisector of the segment joining $z_1$ and $z_2$ , a straight line. The inequality $|z - z_1| < |z - z_2|$ is the half-plane closer to $z_1$ .

Argument conditions give rays

The condition $\arg(z - z_0) = \alpha$ describes all points whose direction from $z_0$ is the fixed angle $\alpha$ . This is a ray (half-line) starting at $z_0$ (open endpoint, since $z = z_0$ has no defined argument) at angle $\alpha$ to the positive real direction. A condition like $0 \leq \arg(z - z_0) \leq \tfrac{\pi}{3}$ is the wedge-shaped region between two rays.

Converting to Cartesian form

To confirm a sketch, substitute $z = x + iy$ . For example $|z - 2| = 3$ becomes $\sqrt{(x-2)^2 + y^2} = 3$ , so $(x - 2)^2 + y^2 = 9$ : a circle, centre $(2, 0)$ , radius $3$ . Squaring both sides removes the surd, turning the modulus condition into a recognisable Cartesian equation.

Boundaries: solid or dashed

Strict inequalities ( $<$ , $>$ ) exclude the boundary, drawn as a dashed curve. Inclusive inequalities ( $\leq$ , $\geq$ ) include the boundary, drawn solid. The marker checks both the correct shape and the correct boundary style, and whether the interior or exterior is shaded.

Combining conditions

Two or more conditions joined by "and" give the intersection of the regions, the overlap. Sketch each region, then shade only the common part. This is where careful boundary handling matters most, since the answer is the set satisfying every condition at once.

Worked example

Sketch the region $\{\, z : |z - 2| \leq 2 \ \text{and}\ \operatorname{Im}(z) \geq 1 \,\}$ .

Interpret the first condition: $|z - 2| \leq 2$ is the closed disc centred at $z_0 = 2$ (the point $(2, 0)$ ) with radius $2$ . The boundary circle $(x - 2)^2 + y^2 = 4$ is solid because the inequality is inclusive.
Interpret the second condition: With $z = x + iy$ , $\operatorname{Im}(z) = y$ , so $y \geq 1$ is the closed half-plane on and above the horizontal line $y = 1$ , drawn solid.
Find the overlap: Shade the part of the disc lying on or above $y = 1$ . The line $y = 1$ cuts the circle where $(x - 2)^2 + 1 = 4$ , so $(x - 2)^2 = 3$ and $x = 2 \pm \sqrt 3$ .
Describe the result: The region is the circular segment of the disc above the chord from $(2 - \sqrt3,\, 1)$ to $(2 + \sqrt3,\, 1)$ , up to the top of the circle at $(2, 2)$ . Both boundaries are solid since both inequalities are inclusive.
Check a point: Take $z = 2 + 1.5i$ : $|z - 2| = 1.5 \leq 2$ holds, and $\operatorname{Im}(z) = 1.5 \geq 1$ holds, so this interior point is correctly in the shaded region.