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VICSpecialist MathematicsSyllabus dot point

How do equations and inequalities involving the modulus and argument of a complex number describe lines, circles, rays and regions on the Argand plane?

Description and sketching of subsets of the complex plane defined by conditions on modulus and argument, including circles zz0=r|z - z_0| = r, perpendicular bisectors za=zb|z - a| = |z - b|, rays arg(zz0)=α\arg(z - z_0) = \alpha, and the regions defined by the corresponding inequalities

A focused answer to the VCE Specialist Mathematics Unit 3 key-knowledge point on subsets of the Argand plane. Circles, perpendicular bisectors, rays and regions defined by modulus and argument conditions, with a verified worked sketch.

Generated by Claude Opus 4.76 min answer

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

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  1. What this dot point is asking
  2. Circles and discs
  3. Perpendicular bisectors
  4. Rays and arguments
  5. Combining conditions
  6. Examples in context
  7. Try this

What this dot point is asking

VCAA wants you to take a condition on a complex number zz, expressed through its modulus or argument, and sketch the set of points it describes on the Argand plane. The standard loci are circles, perpendicular bisectors and rays; inequalities turn these boundaries into regions. The skill is translating algebra into geometry and back.

Circles and discs

The modulus zz0|z - z_0| is the distance from zz to the fixed point z0z_0. Setting that distance equal to a constant gives a circle:

zz0=ris the circle of radius r centred at z0.|z - z_0| = r \quad\text{is the circle of radius } r \text{ centred at } z_0.

The inequalities describe regions: zz0<r|z - z_0| < r is the open interior (disc without boundary), zz0r|z - z_0| \le r is the closed disc, and zz0>r|z - z_0| > r is the exterior. To verify algebraically, write z=x+iyz = x + iy and z0=a+ibz_0 = a + ib; then zz02=(xa)2+(yb)2=r2|z - z_0|^2 = (x - a)^2 + (y - b)^2 = r^2, the Cartesian circle.

Perpendicular bisectors

The condition za=zb|z - a| = |z - b| says zz is equidistant from the fixed points aa and bb. Geometrically this is the perpendicular bisector of the line segment joining aa and bb. The corresponding inequality za<zb|z - a| < |z - b| is the half plane of points closer to aa than to bb.

Rays and arguments

The argument arg(zz0)\arg(z - z_0) is the direction of the displacement from z0z_0 to zz. Fixing it,

arg(zz0)=α\arg(z - z_0) = \alpha

gives a ray: the half line starting at z0z_0 and pointing in the direction α\alpha. The point z0z_0 itself is excluded, since the argument of 00 is undefined; draw an open circle there. A pair of inequalities such as π6arg(zz0)π3\frac{\pi}{6} \le \arg(z - z_0) \le \frac{\pi}{3} describes the angular sector (wedge) between two rays.

Combining conditions

Many questions intersect two or more loci, for example a disc and a sector, and ask for the overlap region. Sketch each locus, shade the region each inequality allows, and take the common part. Mark boundary curves as solid when the inequality is non-strict (\le, \ge) and dashed when strict (<<, >>).

Examples in context

Example 1. z=3|z| = 3 is the circle of radius 33 centred at the origin.

Example 2. z2=z+2i|z - 2| = |z + 2i| is the perpendicular bisector of the segment from 22 to 2i-2i, that is, the points equidistant from (2,0)(2, 0) and (0,2)(0, -2).

Try this

Q1. Describe the locus z4i=5|z - 4i| = 5. [2 marks]

  • Cue. Circle of radius 55 centred at (0,4)(0, 4).

Q2. Sketch arg(z)=3π4\arg(z) = \frac{3\pi}{4}. [2 marks]

  • Cue. Ray from the origin (excluded) into the second quadrant at 135135^\circ.

Q3. Describe the region 1z21 \le |z| \le 2. [2 marks]

  • Cue. The closed annulus between circles of radius 11 and 22 about the origin.

Exam-style practice questions

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

2025 VCAA1 marksSketch { z : z times conjugate(z) = 4, z in C } on the Argand plane.
Show worked answer →

Use the identity z times conjugate(z) = |z|^2 for any complex number z.

So the condition z times conjugate(z) = 4 becomes |z|^2 = 4, that is |z| = 2.

This is the set of all complex numbers whose distance from the origin is 2. The graph is a circle centred at the origin O with radius 2.

For the sketch, draw a full circle of radius 2 about the origin, passing through the points 2, -2, 2i and -2i on the axes.

2023 VCAA1 marksFind the equation of the ray that originates at the real root of z^7 - 1 = 0 and passes through the point represented by cis(2pi/7), in the form Arg(z - z0) = theta, where z0 is in C and theta is measured in radians in terms of pi.
Show worked answer →

The real root of z^7 - 1 = 0 is z = 1, so the ray originates there: z0 = 1.

The direction of the ray is the argument of the vector from z0 = 1 to the point cis(2pi/7) = (cos(2pi/7), sin(2pi/7)).

That vector is (cos(2pi/7) - 1) + i sin(2pi/7). Using cos(t) - 1 = -2sin^2(t/2) and sin(t) = 2sin(t/2)cos(t/2) with t = 2pi/7, the vector is proportional to (-sin(pi/7), cos(pi/7)), whose argument is pi/2 + pi/7.

Hence theta = pi/2 + pi/7 = 9pi/14, and the ray is Arg(z - 1) = 9pi/14.

Geometrically this is the standard result that a chord of the unit circle from angle 0 to angle alpha points at angle pi/2 + alpha/2.