← Complex numbers

8 Loci of complex numbers

A locus equation describes every point z on the Argand plane satisfying some modulus condition. It always looks geometric, but the ONLY reliable way in is algebra: substitute z = x + yi and square. Circle, centre a+bi, radius k |z − (a+bi)| = k ⟺ (x−a)² + (y−b)² = k² Every point a fixed distance k from a fixed complex number. Perpendicular bisector |z − (a+bi)| = |z − (c+di)| ⟺ a straight line Every point equidistant from two fixed complex numbers — the perpendicular bisector of the segment joining them. General locus (ratio of distances) |z − (a+bi)| = k·|z − (c+di)|, k ≠ 1 Squares to a circle if k ≠ 1, and to a straight line if k = 1. Substitute and expand to see which.

The method, every time

  • Substitute z = x + yi into both sides.
  • Square both sides to remove the modulus signs — never try to reason about the shape before doing this.
  • Expand and simplify. The x² and y² terms either cancel (giving a line) or survive with equal coefficients (giving a circle) or unequal coefficients (giving a hyperbola, if it is a DIFFERENCE of distances rather than a ratio).
  • If a line y = mx + c is given as tangent to a circle-locus, substitute it in and set the resulting quadratic's discriminant to zero — that is the Excellence step, not a separate circle-geometry argument.
Worked example A locus given as a ratio of distances The locus of z is described by |z − 1 − 7i| = 2|z − 4 − 4i|. Find the Cartesian equation of the locus, in the form (x−a)² + (y−b)² = k. Let z = x + yi. |(x−1) + (y−7)i| = 2|(x−4) + (y−4)i|. Square both sides: (x−1)² + (y−7)² = 4[(x−4)² + (y−4)²]. Expand: x² −2x +1 +y² −14y +49 = 4x² −32x +16 +4y² −32y +64. Collect: 3x² − 30x + 3y² − 18y + 78 = 0, then divide by 3 and complete the square on both variables. ✓ (x − 5)² + (y − 3)² = 8 — a circle, since the coefficients of x² and y² matched after expanding. ✕

A locus equation is a set of points — convert to x + yi before interpreting it

Something like |z − i| = |z + 1| describes every point equidistant from two fixed points, which IS a straight line (the perpendicular bisector) — but that is only obvious once you substitute z = x + yi and square both sides to clear the modulus signs. Trying to reason about the geometry before doing the algebra is how marks get lost; the algebra tells you what shape you have.

Practise just this