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.

§-Syllabus dot point

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 4.89 min answerUpdated 2026-06-02

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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.

Sketching a circle locus

Describe and sketch the locus $|z - 3 + 2i| = 4$ .

Step 1: Rewrite the modulus in the standard form $|z - a|$

The expression inside the modulus must be forced into the form $z - a$ so we can read off the centre $a$ directly. A sign error here is the most common mistake on locus questions.

Rewrite the inside as $z - (3 - 2i)$ , so this is $|z - (3 - 2i)| = 4$ . This identifies the fixed point as $a = 3 - 2i$ , which sits at $(3, -2)$ on the Argand diagram.

Step 2: Identify the geometric shape and write its Cartesian equation

Since $|z - a| = r$ says the distance from $z$ to the point $a$ is constant, this is by definition a circle. Its centre is at $(3, -2)$ and its radius is $4$ . Writing $z = x + iy$ gives the equivalent Cartesian equation:

(x - 3)^2 + (y + 2)^2 = 16.

Step 3: Sketch the locus

Mark the centre $(3, -2)$ and locate the four cardinal points at distance $4$ in each axis direction to guide the sketch.

To sketch, mark the centre $(3, -2)$ and draw a circle through points such as $(7, -2)$ , $(-1, -2)$ , $(3, 2)$ and $(3, -6)$ .

Final answer: A circle of radius $4$ centred at $(3, -2)$ , with equation $(x - 3)^2 + (y + 2)^2 = 16$ .

A perpendicular bisector and a region

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

Step 1: Identify the boundary and the two fixed points

When two moduli are set equal, the locus is the perpendicular bisector of the segment joining the two fixed points. Writing $|z - 1| = |z - (-i)|$ reveals the fixed points $a = 1$ at $(1, 0)$ and $b = -i$ at $(0, -1)$ .

The midpoint of this segment is $\left(\tfrac12, -\tfrac12\right)$ , and the segment has gradient $\tfrac{-1 - 0}{0 - 1} = 1$ , so the perpendicular bisector has gradient $-1$ and passes through $\left(\tfrac12, -\tfrac12\right)$ .

Step 2: Determine which side of the line satisfies the inequality

The inequality $|z - 1| \le |z + i|$ means $z$ is at most as far from $1$ as from $-i$ , that is, $z$ is at least as close to $1$ as to $-i$ . Geometrically this is the closed half-plane on the side of the perpendicular bisector that contains the point $(1, 0)$ .

Step 3: Confirm algebraically and state the region

Squaring both sides of the boundary condition removes the moduli and gives a linear inequality, which we can verify and use to state the region in Cartesian form.

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 $y \ge -x$ .

Final answer: The region $y \ge -x$ , that is the closed half-plane on and above the line $y = -x$ , represented as a solid line with shading above it.

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.

The weighted-distance circle (an Apollonius circle)

A condition like $|z + 1| = \sqrt{2}\,|z - i|$ , where the two distances are in a fixed ratio other than $1$ , is not a perpendicular bisector. It is still a circle, called an Apollonius circle, and the past question above shows the standard method: write $z = x + iy$ , square both sides to clear the moduli, and complete the square. The ratio $\sqrt{2}$ shifts the centre off the midpoint and sets a finite radius. Recognising that an unequal ratio of moduli gives a circle (while an equal ratio gives a line) saves you from expecting the wrong shape.

Why this matters

Loci questions reward the student who thinks geometrically first and reaches for algebra only to confirm. Knowing the core shapes (circle, perpendicular bisector, ray, Apollonius circle) 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.

TCE 20234 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: $(x + 1)^2 + y^2 = 2\left[x^2 + (y - 1)^2\right]$ .

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.

TCE 20234 marksShow on an Argand diagram the set of points

\{z : |z + 1| \ge \sqrt{2}\,|z - i|\} \cap \{z : \operatorname{Re}(z) \le 0\}

. Include the coordinates of any significant points.

Show worked answer →

Square the first condition: $(x + 1)^2 + y^2 \ge 2\left[x^2 + (y - 1)^2\right]$ , which rearranges to $x^2 - 2x + y^2 - 4y + 1 \le 0$ , that is $(x - 1)^2 + (y - 2)^2 \le 4$ . So the first set is the closed disc centred $(1, 2)$ with radius $2$ .

The second condition $\operatorname{Re}(z) \le 0$ is the closed half-plane $x \le 0$ .

The required region is the overlap: the part of the disc 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 $y = 2 \pm \sqrt{3}$ . The significant points are $(0, 2 + \sqrt{3})$ and $(0, 2 - \sqrt{3})$ . Shade the small circular segment of the disc with $x \le 0$ between these two points, including the boundary arc and the chord on the imaginary axis.