Learn7367Paper 1 · Compulsory contentComplex numbersLoci and regions in the Argand diagram

Complex numbers · Subtopic 5 of 5

Loci and Regions in the Argand Diagram

Read $|z-a|$ as "the distance from $z$ to $a$ " and $\arg(z-a)$ as "the direction from $a$ to $z$ ". Then $|z-a|=k$ is a circle, $|z-a|=|z-b|$ a perpendicular bisector, and $\arg(z-a)=\alpha$ a half-line.

Cambridge 9709 Paper 3Edexcel YFM01 WFM02Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 45 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Interpret $|z-a|$ as a distance and $\arg(z-a)$ as a direction.
Sketch the loci $|z-a|=k$ , $|z-a|=|z-b|$ and $\arg(z-a)=\alpha$ , and find their Cartesian equations.
Shade regions such as $|z-a|\le k$ or $\arg(z-a)<\alpha$ , and combinations of them.
Find the greatest and least values of $|z|$ and $\arg z$ for points on a locus or in a region.

Pause and think

$|z-(3+4i)|$ is the modulus of $z-(3+4i)$ . If $z=x+iy$ , what is it in terms of $x$ and $y$ , and what does it measure on the Argand diagram?

Section 2 of 14

Core idea

Start hereGeometry in disguise

$|z-(2+i)|=2$ looks like algebra, but read it as "the distance from $z$ to the point $2+i$ is $2$ ": a circle. Complex-number conditions are geometric statements, and sketching them is often the quickest way to answer questions about them.

The big idea

$|z-a|$ = distance from $a$ ; $\arg(z-a)$ = direction from $a$ . $|z-a|=r$ : circle. $|z-a|=|z-b|$ : perpendicular bisector. $\arg(z-a)=\theta$ : half-line from $a$ (excluding $a$ ).

1. Why |z − a| is a distance

|z-a|

with

z=x+iy

a=p+iq

$z-a=(x-p)+i(y-q)$
Subtract.
$|z-a|=\sqrt{(x-p)^2+(y-q)^2}$
Modulus.

Exactly the distance formula between

(x,y)

and

(p,q)

|z − (2 + i)| = 2 and the region |z − (2 + i)| ≤ 2

=2

" is the circle itself. "

\le2

" is the circle and everything inside (shaded).

Worked example $|z-2-i|=2$

Describe the locus.

Goal: Shape and Cartesian equation.
Strategy: Rewrite as $|z-(2+i)|$ .

$|z-(2+i)|=2$
Group.
$(x-2)^2+(y-1)^2=4$
Square the distance.

AnswerCircle centre

(2,1)

, radius

2

; "

\le2

" is the inside.

✓ Check

z=4+i

|2|=2

✓ on it.

2. Perpendicular bisectors and half-lines

|z − 1| = |z − 3i|: the perpendicular bisector

Any point

P

on the line is equally far from

1

and from

3i

(the two dashed lines are equal).

Worked example $|z-1|=|z-3i|$

Find the Cartesian equation.

Goal: A line.
Strategy: Equidistant from $(1,0)$ and $(0,3)$ .

$(x-1)^2+y^2=x^2+(y-3)^2$
Square both distances.
$-2x+1=-6y+9$
Expand; $x^2$ , $y^2$ cancel.
$x-3y+4=0$
Rearrange.

Answer

x-3y+4=0

✓ CheckMidpoint

(0.5,1.5)

0.5-4.5+4=0

✓.

arg(z − 1) = π/4 and the region 0 ≤ arg(z − 1) ≤ π/4

The bold ray is

\arg(z-1)=\tfrac\pi4

. The shaded wedge between it and the real axis is

0\le\arg(z-1)\le\tfrac\pi4

Worked exampleLeast and greatest $|z|$ on $|z-3-4i|=2$

Find them.

Goal: Two values.
Strategy: Distance from origin to centre ± radius.

$|3+4i|=5$
Origin to centre.
$5-2=3,\quad5+2=7$
Along the line through the centre.

AnswerLeast

3

, greatest

7

✓ CheckThe origin is outside the circle (

5>2

) ✓.

Pause and think

Describe $|z+2|\le|z-2|$ .

Summary

Circle, perpendicular bisector, half-line.
Use geometry for max/min.

Locus: Set of points satisfying a condition.

HistoryGeometry as algebra. Treating points in the plane as complex numbers let 19th-century mathematicians such as Augustin-Louis Cauchy and Bernhard Riemann turn geometry problems into algebra. Today the whole branch of mathematics built on this, complex analysis, is used in fluid dynamics to design aircraft wings.

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