Loci in the Argand Diagram

A2 · P3 · 4 min

A locus is the set of points $z$ satisfying a condition. In the Argand diagram, $|z - a|$ is the distance from $z$ to the point $a$ , and $\arg(z - a)$ is the direction from $a$ to $z$ . Every locus in the syllabus is a circle, a perpendicular bisector, or a half-line, and inequalities shade one side.

The three standard loci

Key result

Condition	Locus
$\lvert z - a \rvert = k$	circle, centre $a$ , radius $k$
$\lvert z - a \rvert = \lvert z - b \rvert$	perpendicular bisector of the segment joining $a$ and $b$
$\arg(z - a) = \alpha$	half-line from $a$ (not including $a$ ) at angle $\alpha$ to the positive real direction

Inequalities: $\lvert z - a \rvert < k$ is the inside of the circle; $\lvert z - a \rvert > \lvert z - b \rvert$ is the side of the bisector nearer to $b$ ; $\alpha < \arg(z - a) < \beta$ is the wedge between two half-lines.

Write $a$ as a point: $|z - (2 - 3i)|$ measures distance from $(2, -3)$ . Watch the signs: $|z + 1 - 2i| = |z - (-1 + 2i)|$ is distance from $(-1, 2)$ .

Method

Rewrite the condition in the form $|z - a|$ or $\arg(z - a)$ so the point $a$ is visible.
Identify which of the three loci it is and sketch it, marking $a$ and any radius or angle.
For an inequality, shade the correct region and use a dashed line for strict inequalities, solid for $\leq$ or $\geq$ .
For "greatest/least value" questions, use the geometry: the nearest and farthest points of a circle from a point lie on the line through the centre.

A circle

Sketch the locus $|z - 3 + 2i| = 2$ and find the greatest value of $|z|$ on it.

Solution

$|z - (3 - 2i)| = 2$ : circle centre $(3, -2)$ , radius $2$ .

The distance from the origin to the centre is $\sqrt{13}$ , so the greatest $|z|$ is $\sqrt{13} + 2$ (on the far side of the centre) and the least is $\sqrt{13} - 2$ .

A perpendicular bisector

Sketch the locus $|z - 4| = |z + 2i|$ and find its Cartesian equation.

Solution

Points equidistant from $(4, 0)$ and $(0, -2)$ : the perpendicular bisector of the segment joining them.

Algebraically, with $z = x + iy$ : $(x - 4)^2 + y^2 = x^2 + (y + 2)^2 \Rightarrow -8x + 16 = 4y + 4 \Rightarrow y = -2x + 3$ .

A half-line

Sketch the locus $\arg(z + 1) = \tfrac{\pi}{4}$ and find the point on it with $|z|$ least.

Solution

Half-line starting at $(-1, 0)$ going up-right at $45^\circ$ : the line $y = x + 1$ for $x > -1$ .

The closest point to the origin is the foot of the perpendicular from $O$ to the line $y = x + 1$ , which is $\left(-\tfrac{1}{2}, \tfrac{1}{2}\right)$ , i.e. $z = -\tfrac{1}{2} + \tfrac{1}{2}i$ , with $|z| = \tfrac{1}{\sqrt{2}}$ .

A region defined by two conditions

Shade the region where $|z - 2i| \leq 3$ and $0 \leq \arg z \leq \tfrac{\pi}{2}$ , and find the greatest value of $|z|$ in the region.

Solution

The circle has centre $(0, 2)$ and radius $3$ ; the argument condition is the closed first quadrant. The region is the part of the disc lying in the first quadrant, with solid boundaries.

The farthest point of the whole circle from $O$ is directly above the centre, at $(0, 5)$ , which lies on the boundary of the quadrant and so is in the region. The greatest $|z|$ is $5$ .

Greatest and least argument on a circle

Find the greatest and least values of $\arg z$ for points on $|z - 4 - 3i| = 2$ .

Solution

Centre $C(4, 3)$ , $|OC| = 5$ , radius $2$ . The tangents from $O$ to the circle make angle $\beta$ with $OC$ where $\sin\beta = \tfrac{2}{5}$ , so $\beta = 0.4115$ . $\arg$ of the centre is $\tan^{-1}\tfrac{3}{4} = 0.6435$ .

Greatest $\arg z = 0.6435 + 0.4115 = 1.055$ ; least $= 0.6435 - 0.4115 = 0.232$ radians.

Watch out

$\arg(z - a) = \alpha$ is a half-line, not a full line: the points on the other side of $a$ have argument $\alpha \pm \pi$ . Draw an open circle at $a$ itself, because $\arg 0$ is undefined.

Exam tip

Sketches must show: the centre and radius of a circle, the two points and the bisector for a modulus equality, the start point and angle for an argument. Label the axes Re and Im. When the question says "shade the region", make the boundary type (solid or dashed) match the inequality.

Practice

Question

Describe and sketch $|z + 3 - i| = 4$ .
Find the Cartesian equation of $|z - 1| = |z - i|$ .
Sketch $\arg(z - 2) = -\tfrac{\pi}{3}$ .
Find the least value of $|z - 6 - 8i|$ for points satisfying $|z| \leq 3$ .
Shade the region $|z - 1 - i| < 1$ and $\arg z > \tfrac{\pi}{4}$ .

Answers

Circle centre $(-3, 1)$ , radius $4$ .
$y = x$ .
Half-line from $(2, 0)$ downwards to the right at $60^\circ$ below the real axis.
Distance from the origin to $(6, 8)$ is $10$ ; least value $10 - 3 = 7$ .
Inside the circle centre $(1, 1)$ radius $1$ , above the line $y = x$ (dashed boundaries).

The three standard loci

Practice

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