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 , perpendicular bisectors , rays , 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.
Reviewed by: AI editorial process; not yet individually human-reviewed
Have a quick question? Jump to the Q&A page
Jump to a section
What this dot point is asking
VCAA wants you to take a condition on a complex number , 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 is the distance from to the fixed point . Setting that distance equal to a constant gives a circle:
The inequalities describe regions: is the open interior (disc without boundary), is the closed disc, and is the exterior. To verify algebraically, write and ; then , the Cartesian circle.
Perpendicular bisectors
The condition says is equidistant from the fixed points and . Geometrically this is the perpendicular bisector of the line segment joining and . The corresponding inequality is the half plane of points closer to than to .
Rays and arguments
The argument is the direction of the displacement from to . Fixing it,
gives a ray: the half line starting at and pointing in the direction . The point itself is excluded, since the argument of is undefined; draw an open circle there. A pair of inequalities such as 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 (, ) and dashed when strict (, ).
Examples in context
Example 1. is the circle of radius centred at the origin.
Example 2. is the perpendicular bisector of the segment from to , that is, the points equidistant from and .
Try this
Q1. Describe the locus . [2 marks]
- Cue. Circle of radius centred at .
Q2. Sketch . [2 marks]
- Cue. Ray from the origin (excluded) into the second quadrant at .
Q3. Describe the region . [2 marks]
- Cue. The closed annulus between circles of radius and 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.