Argand loci and complex regions: VCE Specialist Mathematics Units 3 & 4

1. A complex condition is a geometric instruction

A locus is a set of points satisfying a condition. In the complex plane, z = x + yi represents the point (x, y), with x and y real. Translating a condition into geometry means identifying all and only the permitted points. A sketch must therefore show more than an approximate shape: it should communicate position, scale, included boundaries and any excluded points.

The most useful interpretation is |z − a| = distance from the point a to the point z. If a = p + qi, then z − a = (x − p) + (y − q)i and |z − a| = √[(x − p)² + (y − q)²]. The subtraction is a displacement from a, not from the origin unless a = 0. This distance interpretation often reveals the locus before any expansion is needed.

There are two complementary methods. Geometry identifies familiar sets quickly: fixed distance gives a circle, equal distances give a perpendicular bisector, and fixed direction gives a ray. Cartesian algebra supplies exact equations, intersections and boundary checks. Use both. An algebraic equation can describe too many points if a sign or argument restriction was lost during conversion.

2. Circles, disks and annuli

For r > 0, |z − a| = r is the circle with centre a and radius r. The condition |z − a| < r is its interior, excluding the circle itself; |z − a| ≤ r includes the boundary and is a closed disk. The inequalities > and ≥ describe the exterior, again with the boundary excluded or included respectively. Use a solid boundary for inclusion and a dashed boundary for exclusion, with a legend if needed.

The exceptional radii are simple but important. The equation |z − a| = 0 gives the single point a. A negative right-hand side gives no points because a modulus cannot be negative. The inequality |z − a| < 0 is also impossible. Check these facts before squaring: squaring an impossible equation such as |z − a| = −2 would incorrectly create a circle.

A double inequality r₁ < |z − a| ≤ r₂, with 0 ≤ r₁ < r₂, gives an annulus between two concentric circles. The inner boundary is excluded and the outer boundary included. Its centre is excluded when r₁ = 0 because the lower inequality is strict. State each boundary separately rather than describing the entire region as simply “inside” or “outside”.

Scale factors can move both the centre and radius. For nonzero complex c, |cz − d| = r becomes |c||z − d/c| = r, so the centre is d/c and the radius is r/|c| when r ≥ 0. In the common real-scale case |2z − 4 + 2i| ≤ 6, factoring 2 gives |z − (2 − i)| ≤ 3. The centre is 2 − i, not 4 − 2i.

3. Equal-distance lines and component conditions

For distinct fixed points a and b, |z − a| = |z − b| describes the perpendicular bisector of the segment joining them. Squaring the two non-negative distances is reversible. The x² and y² terms cancel, leaving a linear equation. If a = b, every point satisfies the equality; the usual perpendicular-bisector description requires distinct points.

Simple component conditions are already Cartesian. Re(z) = c is the vertical line x = c, and Im(z) = d is the horizontal line y = d. Their inequalities give the corresponding half-planes. For example, Im(z) ≥ −2 includes the line y = −2 and all points above it. A condition on the real part is not a condition on the modulus; negative real parts are perfectly possible.

When comparing distances, a boundary test point should not lie on the boundary itself. Substitute a simple point, often one of the two centres, to identify the correct side. If an inequality contains a quotient, preserve any original denominator exclusion before simplifying. A cancelled expression cannot restore points where the original condition was undefined.

4. An ellipse fixes a sum of distances

The condition |z − a| + |z − b| = L fixes the sum of the distances from two points called the foci. When a and b are distinct and L is greater than their separation, the locus is an ellipse. This is assumed Unit 2 complex-number knowledge. A point moves around the curve while its two distances vary, but their total remains constant. Equal distance from the foci would instead give their perpendicular bisector.

For example, |z − 3| + |z + 3| = 10 has foci (3, 0) and (−3, 0). The horizontal vertices are (5, 0) and (−5, 0), since their distance sums are 2 + 8. At x = 0 the two distances are equal, so each is 5; hence 9 + y² = 25 and y = ±4. The Cartesian ellipse is x²/25 + y²/16 = 1. Its unequal axis lengths do not mean the Argand axes use unequal scales.

The inequality |z − 3| + |z + 3| ≤ 10 includes this ellipse and its interior. Check the centre: its distance sum is 6, so it belongs. A total less than the separation of the foci is impossible, because travelling via a third point cannot shorten the straight-line distance between them. A total equal to their separation gives the joining line segment, including its endpoints, rather than a non-degenerate ellipse. If the foci coincide, the sum is twice one distance, so a positive fixed total gives a circle instead.

5. Arguments describe direction, with an excluded origin

For z ≠ a, Arg(z − a) = θ describes the ray beginning at a in direction θ, with a itself removed. The angle is measured anticlockwise from the positive horizontal direction through a. Under the principal convention, −π < θ ≤ π. If θ is outside that interval, an equality involving Arg has no solutions unless the angle is first expressed in the required convention.

The equation Arg(z − a) = π/4, for a = 1 − 2i, gives y + 2 = x − 1 together with x > 1. The linear equation alone would include the opposite ray, whose argument differs by π. The endpoint (1, −2) is excluded because z − a = 0 there and zero has no argument. Draw a hollow point even if the rest of the ray is solid.

Argument inequalities produce sectors. For example, 0 ≤ Arg(z) ≤ π/3 includes the two bounding rays and the points between them, but still excludes the origin. A negative angle represents clockwise rotation. With principal arguments, a sector crossing the negative real axis may need two intervals, such as Arg(z) ≥ 3π/4 or Arg(z) ≤ −3π/4. Do not describe this wraparound sector as one incorrectly ordered interval.

At the negative real axis, the principal argument is π rather than −π. Consequently, interval endpoints require attention. The set −π < Arg(z) < 0 is the open lower half-plane, excluding both real-axis rays and zero. The set 0 ≤ Arg(z) ≤ π includes the upper half-plane and both real-axis rays, with zero removed. The undefined point remains excluded regardless of non-strict angle bounds.

6. Read “and”, “or” and the original domain

Two conditions joined by “and” give an intersection: keep only points satisfying both. Conditions joined by “or” give a union: a point may satisfy either or both. Draw the individual sets lightly before shading the final set. A boundary can be allowed by one condition and removed by another, as in the final worked example.

When technology is permitted, plot the Cartesian equations or inequalities with equal axis scales. Unequal scales can make a circle look elliptical and distort angles. Test representative points inside, outside and on each boundary against the original complex condition. Numerical plotting is particularly unreliable at a deleted point or a strict boundary, so annotate those features explicitly.

For an exact intersection, retain radicals and fractions until the final result. Use decimal coordinates only when requested or as a secondary plotting aid. Squaring a modulus equality is safe when the other side is known non-negative; replacing an argument condition by a tangent equation needs extra directional restrictions. These are different algebraic operations with different risks.

An assessment-ready answer names the shape, centre or line equation, radius or direction, relevant intersections and every inclusion or exclusion. The diagram should have labelled real and imaginary axes and enough exact points to determine its placement. A set written algebraically can support the sketch, but it should not replace an explicitly requested diagram.