CAIE A-Level · Mathematics 9709 · Complex Numbers

Loci in the Argand Diagram (9709 Pure Mathematics 3)

9 min readSyllabus 3.9PreviewBy Uzair Khan

Syllabus objective

Illustrate simple equations and inequalities involving complex numbers by means of loci in an Argand diagram, e.g. |z − a| < k, |z − a| = |z − b|, arg(z − a) = α.

Introduction

Loci in the Argand diagram give a geometric meaning to equations and inequalities involving complex numbers. Rather than solving for a single value of zz, you identify the set of all points z=x+iyz = x + iy that satisfy a given condition. This is examined regularly in 9709 Paper 3: questions may ask you to sketch a locus, describe it in geometric terms, find where two loci intersect, or shade a region defined by an inequality. Mastering these three standard forms — the circle, the perpendicular bisector, and the half-line — is essential.


Core Concept

Every complex number z=x+iyz = x + iy is represented by the point P=(x,y)P = (x, y) in the Argand diagram. The expression zaz - a, where aa is a fixed complex number, represents the displacement vector from the point AA (representing aa) to the point PP (representing zz).

The three loci you need are derived directly from this geometric interpretation:

1. Circle locus: za=k|z - a| = k

za|z - a| is the modulus of zaz - a, i.e. the distance from PP to AA. Setting this equal to a positive constant kk gives all points at a fixed distance kk from AA — a circle with centre AA and radius kk.

The inequality za<k|z - a| < k gives the interior of this circle (excluding the boundary), and zak|z - a| \leq k includes the boundary.

2. Perpendicular bisector locus: za=zb|z - a| = |z - b|

The left side is the distance from PP to AA; the right side is the distance from PP to BB. Equality means PP is equidistant from AA and BB, so the locus is the perpendicular bisector of the line segment ABAB.

3. Half-line locus: arg(za)=α\arg(z - a) = \alpha

arg(za)\arg(z - a) is the argument of the vector from AA to PP, i.e. the angle that AP\overrightarrow{AP} makes with the positive real direction. Fixing this angle at α\alpha gives a half-line (ray) starting at AA (not including AA itself), making angle α\alpha with the positive xx-direction.


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Prerequisites: The Argand Diagram

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