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Loci in the Complex Plane

Key Terms

|z − a| = r
Circle centred at the point representing a, with radius r.
|z − a| = |z − b|
Perpendicular bisector of the line segment joining a and b.
arg(z − a) = θ
Ray starting at a (not including a), making angle θ with the positive real axis.
Re(z) = c
Vertical line x = c.   Im(z) = c: Horizontal line y = c.
Regions
|z − a| < r is the interior (open disk); |z − a| ≤ r includes the boundary (closed disk).
Standard Loci at a Glance:
  • |z| = r  →  circle centre origin, radius r
  • |z − (a + bi)| = r  →  circle centre (a, b), radius r   [Cartesian: (x−a)² + (y−b)² = r²]
  • arg(z) = θ  →  ray from origin at angle θ
  • |z − a| = |z − b|  →  perpendicular bisector of segment from a to b
  • |z − a|/|z − b| = k (k ≠ 1)  →  Apollonius circle
Hot Tip Always write z = x + iy and substitute into the locus equation, then simplify to recognise the geometric shape. For |z − a| = r, expand |x + iy − a| = r by writing a = a1 + ia2 and computing modulus: √[(x−a1)² + (y−a2)²] = r, then square both sides.

Worked Example 1 — Identify the locus |z − 3 + 2i| = 4

Write a = 3 − 2i, so the centre is at (3, −2) and the radius is 4.

Cartesian equation: (x − 3)² + (y + 2)² = 16.

This is a circle, centre (3, −2), radius 4, on the Argand diagram.

Worked Example 2 — Locus |z − 2| = |z + 4|

Write z = x + iy. |z − 2| = |z + 4| means the point is equidistant from 2 (on the real axis) and −4 (on the real axis).

√[(x−2)² + y²] = √[(x+4)² + y²]

Squaring: x² − 4x + 4 + y² = x² + 8x + 16 + y²

−4x + 4 = 8x + 16 ⇒ −12x = 12 ⇒ x = −1.

The locus is the vertical line x = −1, which is the perpendicular bisector of the segment from −4 to 2.

Understanding Loci in the Complex Plane

A locus is the set of all points satisfying a given condition. In the complex plane (Argand diagram), the point z = x + iy corresponds to the Cartesian point (x, y). Locus conditions involving z are translated into geometric conditions on (x, y) by writing z = x + iy and using the definitions of modulus and argument.

The Circle: |z − a| = r

The condition |z − a| = r says: the distance from z to the fixed point a is equal to r. This is the definition of a circle with centre a and radius r.

To find the Cartesian equation, let a = a1 + ia2 and z = x + iy:

√[(x − a1)² + (y − a2)²] = r  ⇒  (x − a1)² + (y − a2)² = r²

Region: |z − a| < r is the open disk (interior of the circle); |z − a| ≤ r is the closed disk (interior plus boundary).

The Ray: arg(z − a) = θ

The condition arg(z − a) = θ says: the angle from the positive real direction to the vector from a to z is θ. This is a ray (half-line) starting at a (but not including a itself), extending in the direction of angle θ.

To write it parametrically: let w = z − a, so arg(w) = θ means w = t(cosθ + i sinθ) for t > 0. Therefore z = a + t(cosθ + i sinθ), t > 0.

In Cartesian form (with a = a1 + ia2): x = a1 + t cosθ, y = a2 + t sinθ, t > 0. Eliminating t: (y − a2)/(x − a1) = tanθ, i.e. y − a2 = tanθ(x − a1), restricted to the ray direction.

The Perpendicular Bisector: |z − a| = |z − b|

This says z is equidistant from two fixed points a and b. The locus is the perpendicular bisector of the segment from a to b. To find its equation, substitute z = x + iy and simplify (the squared terms cancel leaving a linear equation).

Lines: Re(z) = c and Im(z) = c

Re(z) = c means x = c, a vertical line. Im(z) = c means y = c, a horizontal line. More generally, Im(z − a) = 0 (for complex a = a1 + ia2) means the imaginary part of (x − a1) + i(y − a2) is zero, i.e. y = a2.

Apollonius Circles: |z − a|/|z − b| = k

When k ≠ 1, the locus |z − a| = k|z − b| is a circle, known as an Apollonius circle. Substitute z = x + iy, square both sides, and complete the square to find centre and radius.

For example, |z − 1|/|z + 1| = 2 gives |z − 1| = 2|z + 1|, so (x−1)²+y² = 4[(x+1)²+y²]. Expanding: x²−2x+1+y² = 4x²+8x+4+4y². Rearranging: 3x²+10x+3+3y²=0, i.e. x²+(10/3)x+y²=−1. Complete the square: (x+5/3)²+y² = 25/9−1 = 16/9. Centre (−5/3, 0), radius 4/3.

Exam Tip: For any locus question, first identify the type (circle, line, ray) from the form of the equation. Then state the geometric description in words before writing the Cartesian equation. Examiners award marks for correct identification.
Exam Tip: For ray loci, always specify that the endpoint is excluded (open circle on diagrams). arg(z − a) = θ is a ray that starts at a but does NOT include a.

Mastery Practice

  1. Sketch and describe the locus |z − 2| = 3. Fluency

    Write the Cartesian equation of the locus, identify the shape, centre, and radius, and sketch it on an Argand diagram.

  2. Sketch and describe arg(z) = π/4. Fluency

    Identify the locus, write it in Cartesian form, and state any restrictions.

  3. Find the Cartesian equation for |z − 1 − i| = 2. Fluency

    Write the Cartesian equation and state the centre and radius.

  4. Describe and find the Cartesian equation of the locus |z − 3| = |z + 1|. Fluency

    Give the geometric description and derive the equation.

  5. Find and sketch the region |z − 2i| ≤ 3. Understanding

    Describe the region geometrically and write the Cartesian inequality.

  6. Describe and sketch the locus Im(z − 1 + 2i) = 0. Understanding

    Write the Cartesian equation and give a geometric description.

  7. Find all complex numbers satisfying |z| = |z − 4| and Im(z) = 2. Understanding

    Use the locus conditions simultaneously to find z.

  8. Sketch the region {z : |z + i| ≤ 2 and Re(z) ≥ 0}. Understanding

    Describe each condition, find where they overlap, and identify key boundary points.

  9. Write the locus arg(z − 1) = π/3 parametrically and as a Cartesian equation. Problem Solving

    Give the parametric form (x(t), y(t)), convert to Cartesian, and state any domain restrictions.

  10. Apollonius circle: find the centre and radius of {z : |z − 1|/|z + 1| = 2}. Problem Solving

    Show that the locus is a circle, and find its centre and radius in exact form.

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