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Complex loci

Modulus statements describe distances on an Argand diagram.

|z − a| = r describes a circle with centre a and radius r ≥ 0. The equation |z − a| = |z − b| describes points equidistant from a and b: their perpendicular bisector when a ≠ b.

An argument condition usually describes a ray, with the starting point excluded because the argument of zero is undefined. Check boundary inclusion for inequalities. The diagram should show both the geometry and any omitted points.

Worked example

Describe |z − (2 + i)| = 3.

  1. z − (2 + i) is the displacement from (2,1).
  2. Its modulus is the distance from that point.

Answer: Circle with centre (2,1) and radius 3

Revise first: Modulus and argument.

A complex equation can describe a whole line

|z − a| is the distance from z to a on an Argand diagram. The equation |z − a| = |z − b| therefore describes the perpendicular bisector of the segment joining a and b, rather than just its midpoint.

Put z = x + iy and square both sides. Squaring is safe here because distances are nonnegative. The x² and y² terms cancel, leaving a linear equation. Retain that geometric interpretation as a check on your algebra.

Equal distances from 1 + i and 5 + 3i

  1. (x − 1)² + (y − 1)² = (x − 5)² + (y − 3)².
  2. Cancel the squared terms and collect: 8x + 4y = 32, or 2x + y = 8.
  3. The midpoint (3, 2) satisfies the line. Its normal (2, 1) is parallel to the joining vector (4, 2), so the line is perpendicular to the segment.
Does |z − (2 + i)| = 5 imply x² + y² = 25?

No. The circle is centred at (2, 1), so its equation is (x − 2)² + (y − 1)² = 25.

Take the idea further

Intersect that circle with y = 4. The horizontal displacements are ±4, giving x = −2 or 6.

Explore the diagram and test a prediction

More advanced methods

Practise complex loci

Course mapping

These specification references show where the topic occurs. The questions cover only some parts of each topic.

Next practice: Circles and tangents.