What is rays from an argument condition?

The equation (z - a) = θ fixes the direction from the point a to z. The locus is a ray (half-line) starting at a, pointing at angle θ to the positive real axis. The starting point a itself is excluded, because (0) is undefined. A condition like α (z - a) β describes a wedge-shaped region between two rays.

TASSpecialist MathematicsSyllabus dot point

How do equations and inequalities in z describe lines, circles and regions on the Argand diagram?

Sketch curves and regions in the complex plane defined by modulus and argument conditions.

Loci from modulus and argument conditions, circles, perpendicular bisectors, rays and shaded regions on the Argand diagram, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-30

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What this dot point is asking

This dot point asks you to turn algebraic statements about a complex variable $z$ into pictures, and to sketch the set of all points satisfying them. The key translation, developed on the Argand diagram, is that $|z - a|$ is the distance from $z$ to the fixed point $a$ , and $\arg(z - a)$ is the direction from $a$ to $z$ .

Circles from a modulus condition

The equation $|z - a| = r$ says the distance from $z$ to the fixed point $a$ is constant and equal to $r$ . That is exactly the definition of a circle of radius $r$ centred at $a$ . If $a = p + iq$ , writing $z = x + iy$ gives

(x - p)^2 + (y - q)^2 = r^2,

the familiar Cartesian circle. The corresponding region $|z - a| \le r$ is the closed disc, and $|z - a| > r$ is everything outside the circle.

Perpendicular bisectors from equal distances

The equation $|z - a| = |z - b|$ says $z$ is equidistant from the two fixed points $a$ and $b$ . The set of such points is the perpendicular bisector of the segment joining $a$ and $b$ . The inequality $|z - a| < |z - b|$ is the half-plane of points closer to $a$ .

Rays from an argument condition

The equation $\arg(z - a) = \theta$ fixes the direction from the point $a$ to $z$ . The locus is a ray (half-line) starting at $a$ , pointing at angle $\theta$ to the positive real axis. The starting point $a$ itself is excluded, because $\arg(0)$ is undefined. A condition like $\alpha \le \arg(z - a) \le \beta$ describes a wedge-shaped region between two rays.

A perpendicular bisector and a region

Worked example

Sketch the region defined by $|z - 1| \le |z + i|$ .

The points $a = 1$ and $b = -i$ sit at $(1, 0)$ and $(0, -1)$ . The boundary $|z - 1| = |z + i|$ is the perpendicular bisector of the segment joining these two points. Its midpoint is $(\tfrac12, -\tfrac12)$ , and the segment has gradient $\tfrac{-1 - 0}{0 - 1} = 1$ , so the bisector has gradient $-1$ . The inequality $|z - 1| \le |z + i|$ keeps the points at least as close to $1$ as to $-i$ , which is the half-plane containing $(1, 0)$ . Shade that side of the line.

You can confirm algebraically: squaring gives $(x - 1)^2 + y^2 \le x^2 + (y + 1)^2$ , which simplifies to $-2x + 1 \le 2y + 1$ , that is $y \ge -x$ . The origin gives $0 \ge 0$ , true, so the boundary passes through the origin and the region is $y \ge -x$ .

Combining conditions

Many questions intersect two loci, for example a circle and a ray, or shade the overlap of two regions. Sketch each condition separately in light pencil, then identify the common set. Note carefully whether each boundary is included (use a solid line for $\le$ or $\ge$ ) or excluded (a dashed line for strict inequality), and whether endpoints of rays are open circles.

Why this matters

Loci questions reward the student who thinks geometrically first and reaches for algebra only to confirm. Knowing the three core shapes (circle, perpendicular bisector, ray) on sight lets you sketch quickly and spend your time on the intersections and boundary details that earn the final marks.

Exam-style practice questions

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

2023 TASC4 marksShow that the set of points such that |z + 1| = sqrt(2)|z - i| is a circle. State the position of its centre and its radius.

Show worked answer →

Put z = x + iy. Then |z + 1|^2 = (x + 1)^2 + y^2 and |z - i|^2 = x^2 + (y - 1)^2.

Square the given equation to remove the moduli: (x + 1)^2 + y^2 = 2[x^2 + (y - 1)^2].

Expand: x^2 + 2x + 1 + y^2 = 2x^2 + 2y^2 - 4y + 2. Bring everything to one side: x^2 - 2x + y^2 - 4y + 1 = 0.

Complete the square: (x - 1)^2 - 1 + (y - 2)^2 - 4 + 1 = 0, so (x - 1)^2 + (y - 2)^2 = 4. This is a circle with centre (1, 2) and radius 2. Markers reward squaring to clear the moduli and completing the square to read off the centre and radius.

2023 TASC4 marksShow on an Argand diagram the set of points given by {z : |z + 1| >= sqrt(2)|z - i|} intersect {z : Re(z) <= 0}. Include the co-ordinates of any significant points.

Show worked answer →

Square the first condition: (x + 1)^2 + y^2 >= 2[x^2 + (y - 1)^2], which rearranges to x^2 - 2x + y^2 - 4y + 1 <= 0, i.e. (x - 1)^2 + (y - 2)^2 <= 4. So the first set is the closed disk centred (1, 2) with radius 2.

The second condition Re(z) <= 0 is the closed half-plane x <= 0.

The required region is the overlap: the part of the disk lying to the left of (and on) the imaginary axis. Find where the circle meets x = 0: setting x = 0 in x^2 - 2x + y^2 - 4y + 1 = 0 gives y^2 - 4y + 1 = 0, so y = 2 +/- sqrt(3). The significant points are (0, 2 + sqrt(3)) and (0, 2 - sqrt(3)). Shade the small circular segment of the disk with x <= 0 between these two points, including the boundary arc and the chord on the imaginary axis.