How to Solve HSC Extension 2 Complex Number Loci Questions

 

Intro

HSC Extension 2 loci questions ask you to describe or sketch the set of complex numbers satisfying a geometric condition in the Argand plane. The fastest approach is to translate the condition into Cartesian form (or modulus–argument form), identify the shape — circle, line, ray, or half-plane — then sketch with centre, radius, and intercepts labelled. This guide is for NSW Year 12 students preparing for HSC Mathematics Extension 2 under the NESA syllabus. Keywords: complex numbers loci, HSC Ext 2 math tips, Argand diagram questions.

Summary

Loci in Extension 2 are geometry questions written in complex notation: modulus conditions give circles or perpendicular bisectors, argument conditions give rays, and real or imaginary parts give straight lines. Convert to x + iy, simplify, name the shape, then sketch accurately with boundaries stated.

Markers expect precise geometric language: name the centre and radius for circles, state the equation of the line for perpendicular bisectors, and give the direction for rays. Combined loci questions may ask for the area of a region — sketch first, then describe the intersection set.

Key Points

• |z − a| = r describes a circle with centre a and radius r.
• |z − z₁| = |z − z₂| is the perpendicular bisector of the segment joining z₁ and z₂, not a circle.
• Arg(z − a) = θ is a ray from a at angle θ; state whether the endpoint is included.
• Re(z) = k and Im(z) = k are vertical and horizontal lines respectively.
• Always square modulus equations carefully and check for extraneous points lost when squaring.
• Practise with the HSC Complex Numbers booklet on vumaths.com.

Worked example

Question. Find and sketch the locus of z such that |z − 1| = |z + 1|.

Solution.

1. Let z = x + iy. Then |z − 1| = √((x − 1)² + y²) and |z + 1| = √((x + 1)² + y²).
2. Square both sides (non-negative quantities, so equivalence is preserved): (x − 1)² + y² = (x + 1)² + y².
3. Expand: x² − 2x + 1 = x² + 2x + 1, giving −4x = 0, so x = 0.
4. The locus is the imaginary axis — the perpendicular bisector of 1 and −1 on the real axis.
5. Sketch a vertical line through the origin; label that every point on the line is equidistant from 1 and −1.

Answer. The locus is Re(z) = 0 (the imaginary axis).

Takeaway. Equal-modulus conditions to two fixed points always produce a straight-line locus; only a single fixed modulus |z − a| = r gives a circle.

Exam Preparation

Loci appear regularly in Extension 2 papers, often combined with inequalities or regions. In the final month before the HSC, revise the four standard forms, then do timed Argand questions without notes. Keep a one-page summary of modulus, argument, and Re/Im conditions in your error log.

1. Memorise the four locus templates. List circle, perpendicular bisector, ray, and line forms on one revision card.
2. Drill conversion. Do five questions converting complex conditions to Cartesian equations under 15 minutes.
3. Sketch under exam conditions. Include scale, labels, and boundary notes (ray vs full line) as markers expect.

In past HSC Extension 2 papers, loci questions often appear alongside inequalities such as |z − a| ≤ r, where you shade the interior of a circle. When a question combines Re(z) and Im(z) conditions, find the intersection of two lines and state whether the region is bounded. Markers award sketch quality — include scale, labelled axes, and the exact geometric description in words. Pair Argand practice with polynomial root plotting to see the same mathematics from two angles.

Mini-FAQ

Is |z − 1| = |z + 1| a circle?

No. It is the perpendicular bisector of the points 1 and −1 — a straight line. A circle requires a condition like |z − a| = r with one centre and one radius.

When should I use z = re^iθ instead of x + iy?

Use polar form when the condition involves argument or when a rotation is involved. For pure modulus equalities, Cartesian substitution is usually faster.

Do I need to shade regions in loci questions?

If the question asks for a region (e.g. |z − a| ≤ r), shade the interior and state whether the boundary is included. For strict inequalities, use a dashed boundary.

Common mistakes to avoid

• Treating |z − a| = |z − b| as a circle — it is always a line (perpendicular bisector).
• Drawing arg(z) = θ as a full line instead of a ray from the origin (unless stated otherwise).
• Forgetting to label centre and radius when describing |z − a| = r geometrically.
• Mixing degrees and radians when comparing or adding arguments.

Practice on Vu's Maths Hub

Need more practice on this topic? Open the free HSC Complex Numbers booklet on Vu's Maths Hub — worked examples and exam-style questions, readable in your browser with no account required.

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