Loci in the Argand Diagram

A locus is the set of all points $z$z in the Argand diagram satisfying a stated condition. The whole subject rests on one translation: $|z - z_1|$∣z−z1​∣ means the distance from the point $z$z to the point $z_1$z1​, and $\arg(z - z_1)$arg(z−z1​) means the direction of $z$z as seen from $z_1$z1​. Read every locus condition through those two phrases and the geometry — circles, perpendicular bisectors, half-lines, arcs — appears almost automatically. This topic is where the algebra of complex numbers meets the geometry of the plane, and a clean, labelled sketch is worth as many marks as the algebra that justifies it.

---

Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2, building directly on modulus and argument. It is unusually balanced across the assessment objectives: AO1 (convert a locus to a Cartesian equation), AO2 (recognise and justify the geometric shape, and reason about regions and intersections) and AO3 (combine two loci, find intersection points, or interpret a real-world-flavoured constraint). Examiners frequently set a multi-part question that asks you to sketch one or two loci, give a Cartesian equation, and then find where they meet — so fluency in all four standard locus types, and in moving between the complex and Cartesian descriptions, is essential.

---

Core theory: the two readings and the four standard loci

Throughout, write $z = x + iy$z=x+iy so that $|z - z_1| = \sqrt{(x - a)^2 + (y - b)^2}$∣z−z1​∣=(x−a)2+(y−b)2​ when $z_1 = a + bi$z1​=a+bi. Everything follows from the two readings:

• $|z - z_1| =$∣z−z1​∣= distance from $z$z to $z_1$z1​;
• $\arg(z - z_1) =$arg(z−z1​)= angle the line from $z_1$z1​ to $z$z makes with the positive real direction.

Type 1 — a circle: $|z - z_1| = r$∣z−z1​∣=r

"All points a fixed distance $r$r from $z_1$z1​" is, by definition, a circle of radius $r$r centred at $z_1$z1​. Substituting $z = x + iy$z=x+iy and $z_1 = a + bi$z1​=a+bi and squaring gives the Cartesian form

$(x - a)^2 + (y - b)^2 = r^2.$(x−a)2+(y−b)2=r2.

The strict inequalities give regions: $|z - z_1| < r$∣z−z1​∣<r is the open interior (disc without boundary) and $|z - z_1| > r$∣z−z1​∣>r the exterior.

Type 2 — a perpendicular bisector: $|z - z_1| = |z - z_2|$∣z−z1​∣=∣z−z2​∣

"Equidistant from two fixed points" is the perpendicular bisector of the segment joining $z_1$z1​ and $z_2$z2​. It is always a straight line. The inequality $|z - z_1| < |z - z_2|$∣z−z1​∣<∣z−z2​∣ is the half-plane on the $z_1$z1​-side of that bisector.

Type 3 — a half-line: $\arg(z - z_1) = \theta$arg(z−z1​)=θ

Fixing the direction of $z$z from $z_1$z1​ gives a half-line (ray) starting at $z_1$z1​ and going off at angle $\theta$θ to the positive real axis. Two cautions that earn marks:

• it is a half-line, not a full line — only the single direction $\theta$θ qualifies (the opposite direction $\theta + \pi$θ+π does not);
• the start point $z_1$z1​ is excluded, because $\arg(0)$arg(0) is undefined. Mark it with an open circle.

Type 4 — an arc of a circle: $\arg\!\left(\dfrac{z - z_1}{z - z_2}\right) = \theta$arg(z−z2​z−z1​​)=θ

Since $\arg\!\left(\frac{z - z_1}{z - z_2}\right) = \arg(z - z_1) - \arg(z - z_2)$arg(z−z2​z−z1​​)=arg(z−z1​)−arg(z−z2​), this fixes the angle subtended at $z$z by the chord from $z_1$z1​ to $z_2$z2​. By the inscribed-angle theorem, the points from which a fixed chord subtends a constant angle form an arc of a circle through $z_1$z1​ and $z_2$z2​. The important special case is $\theta = \frac{\pi}{2}$θ=2π​: the angle in a semicircle is a right angle, so the locus is a semicircle with $[z_1 z_2]$[z1​z2​] as diameter (endpoints excluded, as the fraction is undefined there).

Condition	Locus	Cartesian type
$\lvert z - z_1\rvert = r$∣z−z1​∣=r	circle, centre $z_1$z1​, radius $r$r	$(x-a)^2 + (y-b)^2 = r^2$(x−a)2+(y−b)2=r2
$\lvert z - z_1\rvert = \lvert z - z_2\rvert$∣z−z1​∣=∣z−z2​∣	perpendicular bisector of $[z_1 z_2]$[z1​z2​]	straight line
$\arg(z - z_1) = \theta$arg(z−z1​)=θ	half-line from $z_1$z1​ (excluded) at angle $\theta$θ	ray
$\arg\!\left(\frac{z - z_1}{z - z_2}\right) = \theta$arg(z−z2​z−z1​​)=θ	arc through $z_1, z_2$z1​,z2​ ($=$= semicircle if $\theta = \frac{\pi}{2}$θ=2π​)	arc of a circle

---

Worked examples (with mark scheme)

Example 1 — a circle and its Cartesian form

Sketch the locus $|z - 3 - 2i| = 4$∣z−3−2i∣=4 and give its Cartesian equation.

Rewrite the condition with a single subtraction: $|z - (3 + 2i)| = 4$∣z−(3+2i)∣=4. This is the set of points at distance $4$4 from $3 + 2i$3+2i — a circle, centre $(3, 2)$(3,2), radius $4$4. (M1 identify centre and radius.) Substituting $z = x + iy$z=x+iy:

$(x - 3)^2 + (y - 2)^2 = 16. \quad (\text{A1})$(x−3)2+(y−2)2=16.(A1)

(M1 for reading off centre $(3,2)$(3,2) and radius $4$4; A1 for the Cartesian equation. Sketch: a circle through $(7,2)$(7,2), $(-1,2)$(−1,2), $(3,6)$(3,6), $(3,-2)$(3,−2), with the centre marked.)

Example 2 — a perpendicular bisector

Find the Cartesian equation of the locus $|z - 2| = |z - 4i|$∣z−2∣=∣z−4i∣ and interpret it.

Let $z = x + iy$z=x+iy. The condition is "equidistant from $2$2 and $4i$4i":

$\sqrt{(x - 2)^2 + y^2} = \sqrt{x^2 + (y - 4)^2}. \quad (\text{M1 set up both distances})$(x−2)2+y2​=x2+(y−4)2​.(M1 set up both distances)

Square both sides and expand:

$(x - 2)^2 + y^2 = x^2 + (y - 4)^2 \;\Longrightarrow\; x^2 - 4x + 4 + y^2 = x^2 + y^2 - 8y + 16. \quad (\text{M1 square and expand})$(x−2)2+y2=x2+(y−4)2⟹x2−4x+4+y2=x2+y2−8y+16.(M1 square and expand)

The $x^2$x2 and $y^2$y2 terms cancel, leaving $-4x + 4 = -8y + 16$−4x+4=−8y+16, i.e.

$8y - 4x = 12 \;\Longrightarrow\; 2y - x = 3 \;\Longrightarrow\; y = \frac{x + 3}{2}. \quad (\text{A1})$8y−4x=12⟹2y−x=3⟹y=2x+3​.(A1)

This straight line is the perpendicular bisector of the segment from $2$2 (i.e. $(2,0)$(2,0)) to $4i$4i (i.e. $(0,4)$(0,4)). (M1 setting equal distances; M1 squaring; A1 the line. The cancellation of the quadratic terms is the signature of a bisector — if they did not cancel you would have made a slip.)

Example 3 — a half-line

Sketch the locus $\arg(z - 1 - i) = \dfrac{\pi}{4}$arg(z−1−i)=4π​.

Write it as $\arg(z - (1 + i)) = \frac{\pi}{4}$arg(z−(1+i))=4π​. This is a half-line starting at $(1, 1)$(1,1), going off at $\frac{\pi}{4}$4π​ (i.e. $45^\circ$45∘) to the positive real direction — the ray heading up and to the right. (M1 start point; A1 direction with the start excluded.) The start point $(1,1)$(1,1) is not included (open circle), since $\arg(0)$arg(0) is undefined; the ray in the opposite direction ($\arg = -\frac{3\pi}{4}$arg=−43π​) is not part of the locus.

Example 4 — a semicircular arc

Sketch the locus $\arg\!\left(\dfrac{z - 2}{z + 2}\right) = \dfrac{\pi}{2}$arg(z+2z−2​)=2π​.

Here $z_1 = 2$z1​=2 and $z_2 = -2$z2​=−2. The condition fixes the angle subtended by the chord from $2$2 to $-2$−2 at $\frac{\pi}{2}$2π​ — a right angle — so the locus is the semicircle on $[-2, 2]$[−2,2] as diameter. (M1 recognise the right-angle/semicircle case.) The chord midpoint is the origin and its half-length is $2$2, so the circle has centre $(0,0)$(0,0) and radius $2$2. Because the argument is $+\frac{\pi}{2}$+2π​ (positive), it is the upper semicircle; the endpoints $2$2 and $-2$−2 are excluded (the fraction is undefined there). (A1 centre, radius, correct half, endpoints excluded.)

Example 5 — a region from two conditions

Describe and shade the region satisfying both $|z - 2i| \le 3$∣z−2i∣≤3 and $\operatorname{Re}(z) \ge 0$Re(z)≥0.

The first condition is the closed disc of radius $3$3 centred at $(0, 2)$(0,2) (boundary solid, since $\le$≤). The second, $\operatorname{Re}(z) \ge 0$Re(z)≥0, is the right half-plane $x \ge 0$x≥0. Their intersection is the part of that disc lying on or to the right of the imaginary axis — the right-hand portion of the disc, bounded by the arc and a vertical chord along $x = 0$x=0. (M1 each region correctly identified; A1 the correct overlap, boundaries solid.)

Example 6 — circle and Cartesian form, the other way

Find the Cartesian equation of $|z + 1 - 3i| = 5$∣z+1−3i∣=5, and state the centre and radius.

Rewrite as a single subtraction: $|z - (-1 + 3i)| = 5$∣z−(−1+3i)∣=5. This is the circle of radius $5$5 centred at $(-1, 3)$(−1,3). (M1 read off centre and radius.) With $z = x + iy$z=x+iy,

$(x + 1)^2 + (y - 3)^2 = 25. \quad (\text{A1})$(x+1)2+(y−3)2=25.(A1)

(M1 for the centre $(-1, 3)$(−1,3) — note the sign flips from the $+1 - 3i$+1−3i in the condition; A1 for the Cartesian equation. The most common slip here is reading the centre as $(1, -3)$(1,−3): always convert to $|z - z_1|$∣z−z1​∣ form before reading off $z_1$z1​.)

Example 7 — an Apollonius circle

Find the Cartesian equation of the locus $|z - 4| = 2|z - 1|$∣z−4∣=2∣z−1∣, and identify the curve.

Let $z = x + iy$z=x+iy. Squaring $|z - 4| = 2|z - 1|$∣z−4∣=2∣z−1∣ (a ratio of distances, so not a bisector):

$(x - 4)^2 + y^2 = 4\big((x - 1)^2 + y^2\big). \quad (\text{M1 square both sides, with the factor } 4)$(x−4)2+y2=4((x−1)2+y2).(M1 square both sides, with the factor 4)

Expand: $x^2 - 8x + 16 + y^2 = 4x^2 - 8x + 4 + 4y^2$x2−8x+16+y2=4x2−8x+4+4y2. The $-8x$−8x terms cancel, giving $x^2 + 16 + y^2 = 4x^2 + 4 + 4y^2$x2+16+y2=4x2+4+4y2; collecting, $3x^2 + 3y^2 = 12$3x2+3y2=12, so

$x^2 + y^2 = 4. \quad (\text{A1})$x2+y2=4.(A1)

This is a circle, centre the origin, radius $2$2 — an Apollonius circle (the locus of points whose distances to two fixed points are in a fixed ratio $k \neq 1$k=1). (M1 squaring with the squared ratio; A1 the circle. The signature here is that the $x^2, y^2$x2,y2 terms do not fully cancel — unlike a bisector — leaving a genuine circle.)

---

Specimen-style exam question

(specimen-style — not from any past paper)

On a single Argand diagram, sketch the locus $C_1: |z - 4 - 3i| = 5$C1​:∣z−4−3i∣=5 and the locus $C_2: |z| = |z - 8|$C2​:∣z∣=∣z−8∣. (a) Give the centre and radius of $C_1$C1​ and the Cartesian equation of $C_2$C2​. (b) Find the coordinates of the points where $C_1$C1​ and $C_2$C2​ intersect.

(a) $C_1$C1​ is a circle, centre $(4, 3)$(4,3), radius $5$5: $(x-4)^2 + (y-3)^2 = 25$(x−4)2+(y−3)2=25. For $C_2$C2​, points equidistant from $0$0 and $8$8:

$x^2 + y^2 = (x - 8)^2 + y^2 \;\Longrightarrow\; 0 = -16x + 64 \;\Longrightarrow\; x = 4,$x2+y2=(x−8)2+y2⟹0=−16x+64⟹x=4,

the vertical perpendicular bisector of the segment from $0$0 to $8$8.

(b) Substitute $x = 4$x=4 into the circle:

$(4 - 4)^2 + (y - 3)^2 = 25 \;\Longrightarrow\; (y - 3)^2 = 25 \;\Longrightarrow\; y - 3 = \pm 5 \;\Longrightarrow\; y = 8 \text{ or } y = -2.$(4−4)2+(y−3)2=25⟹(y−3)2=25⟹y−3=±5⟹y=8 or y=−2.