Loci in the Argand Diagram

Loci in the Argand Diagram

A locus (plural: loci) is the set of all points z in the Argand diagram that satisfy a given condition. Loci questions combine geometric intuition with algebraic manipulation of complex numbers.

Key Notation

Let z = x + iy represent a general point, and let z₁, z₂ be fixed complex numbers.

Symbol	Meaning
\	z − z₁\		Distance from z to the point z₁
arg(z − z₁)	Angle from z₁ to z, measured from the positive real direction
Re(z)	Real part x
Im(z)	Imaginary part y

Locus Type 1: |z − z₁| = r (Circle)

The set of points at distance r from z₁ is a circle, centre z₁, radius r.

Example: |z − (3 + 2i)| = 5 is a circle centred at (3, 2) with radius 5.

In Cartesian form: (x − 3)² + (y − 2)² = 25.

Special cases:

• |z| = r → circle centred at the origin, radius r
• |z − z₁| < r → interior of the circle (open disc)
• |z − z₁| ≤ r → closed disc including the boundary

Locus Type 2: |z − z₁| = |z − z₂| (Perpendicular Bisector)

The set of points equidistant from z₁ and z₂ is the perpendicular bisector of the line segment from z₁ to z₂.

Example: |z − 2| = |z − 4i|

• Squaring: (x − 2)² + y² = x² + (y − 4)²
• Expanding: x² − 4x + 4 + y² = x² + y² − 8y + 16
• Simplifying: −4x + 4 = −8y + 16, so y = (1/2)x + 3/2

This is the perpendicular bisector of the segment joining (2, 0) and (0, 4).

Locus Type 3: arg(z − z₁) = α (Half-Line)

The set of points where the angle from z₁ to z equals α is a half-line (ray) starting at z₁.

Important: This is a half-line, not a full line. The point z₁ itself is not included (shown as an open circle).

Example: arg(z − (1 + i)) = π/4 is a half-line from (1, 1) making an angle of π/4 with the positive real direction (so going in the direction "north-east").

The gradient of the half-line is tan α.

Locus Type 4: arg((z − z₁)/(z − z₂)) = α (Arc of a Circle)

This is the most challenging locus. The condition arg((z − z₁)/(z − z₂)) = α defines a major or minor arc of a circle through z₁ and z₂.

Key geometric interpretation: The angle subtended at z by the chord from z₁ to z₂ is constant and equals α.

Special cases:

• If α = π/2, the locus is a semicircle with diameter z₁z₂ (the angle in a semicircle is 90°)
• If α = π, the locus is the line segment from z₁ to z₂ (excluding endpoints)
• Acute α → major arc; obtuse α → minor arc

Method to find the circle:

1. Let z = x + iy and write z − z₁ and z − z₂

2. Use arg(w₁/w₂) = arg(w₁) − arg(w₂)

3. Apply tan to both sides and use tan(A − B) formula

4. Rearrange to get a circle equation

Locus Type 5: |z − z₁|/|z − z₂| = k (Apollonius Circle)

When k ≠ 1, this gives the Apollonius circle — the locus of points whose distances from z₁ and z₂ are in the constant ratio k.

Method: Set |z − z₁|² = k²|z − z₂|², expand, and collect terms.

Example: |z − 2|/|z + 2| = 2

• |z − 2|² = 4|z + 2|²
• (x−2)² + y² = 4[(x+2)² + y²]
• x² − 4x + 4 + y² = 4x² + 16x + 16 + 4y²
• 3x² + 20x + 3y² + 12 = 0
• (x + 10/3)² + y² = 64/9
• Circle centred at (−10/3, 0), radius 8/3.

When k = 1, you get the perpendicular bisector (Locus Type 2).

Intersections and Regions

Exam questions often ask you to find intersections or shade regions.

Finding intersections:

• Solve the equations of two loci simultaneously
• For a circle and a line, substitute the line into the circle equation

Shading regions:

• |z − z₁| < r → shade the interior of the circle
• |z − z₁| < |z − z₂| → shade the side of the perpendicular bisector closer to z₁
• 0 < arg(z − z₁) < α → shade the region between two half-lines from z₁

Minimum and Maximum |z| Problems

These ask for the closest and farthest points on a locus from the origin.

For a circle centred at c with radius r:

• Minimum |z| = |c| − r (if the origin is outside the circle)
• Maximum |z| = |c| + r
• These occur along the line from the origin through the centre

For a half-line:

• The minimum |z| is the perpendicular distance from the origin to the line containing the half-line (but only if the foot of the perpendicular lies on the ray, not its extension behind z₁)

Worked Example

Find the locus of points satisfying both |z − 4| = |z − 2i| and |z| ≤ 5.

Step 1: |z − 4| = |z − 2i| gives the perpendicular bisector of (4, 0) and (0, 2).

• (x−4)² + y² = x² + (y−2)²
• −8x + 16 = −4y + 4
• y = 2x − 3

Step 2: |z| ≤ 5 is the disc x² + y² ≤ 25.

Step 3: The locus is the line segment of y = 2x − 3 that lies inside the circle x² + y² = 25.

Substituting: x² + (2x−3)² = 25 → 5x² − 12x − 16 = 0 → x = (12 ± √(144 + 320))/10 = (12 ± √464)/10.

Exam Tips

• Always sketch the locus — the diagram often reveals the answer faster than algebra
• For arg loci, remember: half-line, not full line, and mark the open circle at the starting point
• When the question says "shade the region", use a test point (often z = 0) to check which side to shade
• For min/max |z| on a circle, draw a line from the origin through the centre and find where it meets the circle
• The arg of a quotient arg(z − z₁)/(z − z₂) = arg(z − z₁) − arg(z − z₂) — but watch for the principal argument restriction −π < arg ≤ π