Loci in the Argand Diagram

This lesson covers loci — sets of points in the Argand diagram that satisfy a given condition involving a complex variable $z$z. Loci problems are a key part of AQA Further Mathematics and combine algebraic and geometric reasoning.

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Locus Type 1: $|z - z_1| = r$∣z−z1​∣=r (Circle)

The set of all points $z$z satisfying:

$|z - z_1| = r$∣z−z1​∣=r

is a circle with centre $z_1$z1​ and radius $r$r.

Reasoning: $|z - z_1|$∣z−z1​∣ is the distance from $z$z to $z_1$z1​, so the locus is all points at distance $r$r from $z_1$z1​.

Worked Example 1: Sketch the locus $|z - 3 - 2i| = 4$∣z−3−2i∣=4.

Rewrite as $|z - (3 + 2i)| = 4$∣z−(3+2i)∣=4.

This is a circle centred at $(3, 2)$(3,2) with radius 4.

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Locus Type 2: $|z - z_1| = |z - z_2|$∣z−z1​∣=∣z−z2​∣ (Perpendicular Bisector)

The set of all points $z$z satisfying:

$|z - z_1| = |z - z_2|$∣z−z1​∣=∣z−z2​∣

is the perpendicular bisector of the line segment joining $z_1$z1​ and $z_2$z2​.

Reasoning: The points equidistant from $z_1$z1​ and $z_2$z2​ form the perpendicular bisector.

Worked Example 2: Find the Cartesian equation of the locus $|z - 2| = |z - 4i|$∣z−2∣=∣z−4i∣.

Let $z = x + yi$z=x+yi.

$|x + yi - 2| = |x + yi - 4i|$∣x+yi−2∣=∣x+yi−4i∣ $\sqrt{(x - 2)^2 + y^2} = \sqrt{x^2 + (y - 4)^2}$(x−2)2+y2​=x2+(y−4)2​

Squaring both sides:

$(x - 2)^2 + y^2 = x^2 + (y - 4)^2$(x−2)2+y2=x2+(y−4)2 $x^2 - 4x + 4 + y^2 = x^2 + y^2 - 8y + 16$x2−4x+4+y2=x2+y2−8y+16 $-4x + 4 = -8y + 16$−4x+4=−8y+16 $8y - 4x = 12$8y−4x=12 $2y - x = 3 \quad \text{or} \quad y = \frac{x + 3}{2}$2y−x=3ory=2x+3​

This is a straight line — the perpendicular bisector of the segment from $2$2 to $4i$4i.

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Locus Type 3: $\arg(z - z_1) = \theta$arg(z−z1​)=θ (Half-line)

The set of all points $z$z satisfying:

$\arg(z - z_1) = \theta$arg(z−z1​)=θ

is a half-line (or ray) starting at $z_1$z1​ at angle $\theta$θ to the positive real direction.

Important: This is a half-line, not a full line. The point $z_1$z1​ itself is not included (since $\arg(0)$arg(0) is undefined).

Worked Example 3: Sketch the locus $\arg(z - 1 - i) = \frac{\pi}{4}$arg(z−1−i)=4π​.

This is a half-line starting at $(1, 1)$(1,1) making an angle of $\frac{\pi}{4}$4π​ (45°) with the positive real direction. It is the ray going to the upper right from $(1, 1)$(1,1), with $(1, 1)$(1,1) not included (shown as an open circle).

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Locus Type 4: $\arg\left(\frac{z - z_1}{z - z_2}\right) = \theta$arg(z−z2​z−z1​​)=θ (Arc of a Circle)

The set of all points $z$z satisfying:

$\arg\left(\frac{z - z_1}{z - z_2}\right) = \theta$arg(z−z2​z−z1​​)=θ

is an arc of a circle passing through $z_1$z1​ and $z_2$z2​.

The angle $\theta$θ is the angle subtended at $z$z by the chord from $z_1$z1​ to $z_2$z2​.

Special case: If $\theta = \frac{\pi}{2}$θ=2π​, then the locus is a semicircle with diameter from $z_1$z1​ to $z_2$z2​ (the angle in a semicircle is $\frac{\pi}{2}$2π​).

Worked Example 4: Sketch the locus $\arg\left(\frac{z - 2}{z + 2}\right) = \frac{\pi}{2}$arg(z+2z−2​)=2π​.

Here $z_1 = 2$z1​=2 and $z_2 = -2$z2​=−2. The locus is a semicircle with diameter from $(-2, 0)$(−2,0) to $(2, 0)$(2,0), specifically the upper semicircle (since the argument is positive $\frac{\pi}{2}$2π​). The points $z_1 = 2$z1​=2 and $z_2 = -2$z2​=−2 are not included.