Loci in the Argand diagram

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An equation or inequality in a complex number z describes a set of points in the Argand diagram, called a locus. Reading |z-a| as the distance from z to a, and arg(z-a) as the direction from a to z, turns these conditions into circles, perpendicular bisectors and half-lines. This lesson sketches those loci, combines them into regions, and uses the geometry to find the greatest and least values of a modulus or an argument without calculus.

Read each modulus as a distance

If the points Z and C represent z and c, then z-c is the displacement from C to Z, so |z-c| is the distance CZ. A locus is the set of all points z that satisfy a condition. Reading each modulus as a distance turns the condition into geometry:

Condition	Locus
\|z-c\| = r	circle, centre c, radius r
\|z-c\| ≤ r	the disc: the circle and its inside
\|z-c\| < r	the inside only; draw the circle dashed
\|z-a\| = \|z-b\|	perpendicular bisector of the points a and b
arg(z-a) = α	half-line from a at angle α, not including a

Watch the sign inside the modulus: |z + 2i| = |z-(-2i)| measures the distance from -2i, not from 2i.

Describe the locus |z-3-4i| = 5. Find the greatest and least values of |z| for points on it.

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Name the circle
Method
Write the condition as |z-(3 + 4i)| = 5.
Reason
The modulus is the distance from the fixed point 3 + 4i.
Working
Answer
The locus is the circle with centre 3 + 4i and radius 5.
Use the line through the origin and the centre
Method
|z| is the distance from the origin to z.
Reason
The nearest and farthest points of a circle from any point lie on the line through that point and the centre.
Working
Answer
The origin is |3 + 4i| = 5 from the centre, exactly the radius, so the circle passes through the origin. The least value of |z| is 0, and the greatest is 5 + 5 = 10, at 6 + 8i.

Quick check

Check: |6 + 8i-3-4i| = |3 + 4i| = 5, so 6 + 8i is on the circle, and |6 + 8i| = 10. If the origin were inside the circle, the least value would be r minus the distance to the centre instead.

Describe the locus |z-2| = |z + 2i| in words, and find its Cartesian equation.

The geometric description and the Cartesian equation

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It is the set of points equidistant from 2 and -2i: the perpendicular bisector of the segment joining (2,0) and (0,-2).

With z = x + iy, (x-2)² + y² = x² + (y + 2)², which simplifies to -4x + 4 = 4y + 4, so y = -x. Check: the midpoint (1,-1) lies on y = -x, and the line’s gradient -1 is perpendicular to the segment’s gradient 1.

The complex number z satisfies |z-3-4i| = 2. Find the least and greatest values of arg z, in radians to three decimal places.

Least arg z, 3 decimal places

Greatest arg z, 3 decimal places

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The locus is a circle with centre 3 + 4i and radius 2, at distance 5 from the origin. The extreme arguments occur on the two tangents from O, each making an angle α with the line to the centre, where sin α = 2/5:

arg z min = tan⁻¹ 4/3- sin⁻¹ 2/5 ≈ 0.927-0.412 = 0.516,; arg z max = tan⁻¹ 4/3 + sin⁻¹ 2/5 ≈ 1.339.

The tangent is perpendicular to the radius, which is why the angle uses sin with the hypotenuse 5.

An argument fixes a direction from a point

arg(z-a) = α says that the displacement from a to z points in direction α. The locus is a half-line starting at a. The point a itself is excluded, because z-a = 0 has no argument.

For example, arg(z-1-i) = π/4 is the part of the line y = x with x > 1. It meets the circle |z| = 2 where x = y and x² + y² = 4, so x = ± square root of 2. Only x = square root of 2 satisfies x > 1, so there is one intersection, square root of 2 + square root of 2 i. The other solution, - square root of 2 - square root of 2 i, is on the full line but on the wrong side of 1 + i.

Combining conditions into a region

Two conditions together

The disc |z − 2 − 2i| ⩽ 2 cut by the half-plane Re z ⩽ 2.

|z| ⩽ |z − 4| keeps the points no farther from 0 than from 4: the half-plane Re z ⩽ 2. Together with the disc this leaves the shaded half-disc.

Figure data

This table is included in print while it is open.

Key points of the region
Point	Coordinates	Role
2 + 2i	(2, 2)	Centre of the circle
2	(2, 0)	Lower end of the straight edge
2 + 4i	(2, 4)	Upper end of the straight edge
2i	(0, 2)	Leftmost point of the disc

Common mistakes

Reading |z + 2-i| as a distance from 2-i; it is the distance from -2 + i.
Drawing arg(z-a) = θ as a whole line instead of a half-line starting at, and excluding, a.
Shading the wrong side of a perpendicular bisector; test one point, such as the origin.
Taking an extreme modulus from the centre of a circle and forgetting to add or subtract the radius.
Using solid boundaries for strict inequalities, or including the excluded starting point of a half-line.

Before you move on

The region R is given by |z-2-2i| ≤ 2 and |z| ≤ |z-4|. Sketch R and find the greatest value of |z| for z in R, with the point where it occurs.

Write your answer, then compare it with the solution

Your description of R, the greatest modulus and where it occurs

Hint 1

Start from the whole disc

Without the second condition, the farthest point would be on the line from the origin through the centre. Check whether that point is still in R.

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Identify the region
Method
The first condition is the disc with centre 2 + 2i and radius 2; the second is Rez ≤ 2.
Reason
|z| ≤ |z-4| means z is no farther from 0 than from 4, the side of the bisector Rez = 2 containing 0.
Working
Answer
R is the left half of the disc, including both boundaries.
Test the unconstrained farthest point
Method
For the whole disc, the farthest point is (2 + square root of 2)(1 + i).
Reason
Its real part 2 + square root of 2 is greater than 2, so it is not in R.
Working
Answer
On the remaining arc, write z = 2 + 2 cos t + i(2 + 2 sin t) with cos t ≤ 0. Then
|z|² = 12 + 8(cos t + sin t),
which is greatest on that arc when cos t = 0 and sin t = 1, giving |z|² = 20. On the straight edge Rez = 2, |z| also grows with the imaginary part, up to 2 + 4i.
State the answer
Working
Answer
The greatest value of |z| is square root of 20 = 2 square root of 5, at z = 2 + 4i.

Quick check

2 + 4i satisfies both conditions with equality: |2i| = 2 and |2 + 4i| = |-2 + 4i|. A region’s extreme often moves to a corner when a second condition cuts off the unconstrained answer. Compare your sketch and reasoning with the solution; this answer is self-checked, not marked.

The excluded loci |z-a| = k|z-b| with k ≠ 1 and arg(z-a)- arg(z-b) = α are outside this syllabus.

Syllabus and review details

GCE A-Level H2 Further Mathematics 2027 · 2027

Last reviewed: 2026-11-06

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Guided steps

Reading settings

Read each modulus as a distance

Nearest and farthest points of a circle

Name the circle

Use the line through the origin and the centre

Equal distances

The steepest and shallowest directions

An argument fixes a direction from a point

Combining conditions into a region

Common mistakes

Before you move on

The farthest point of a half-disc

Identify the region

Test the unconstrained farthest point

State the answer

Lesson done

Read each modulus as a distance

Nearest and farthest points of a circle

Equal distances

The steepest and shallowest directions

An argument fixes a direction from a point

Combining conditions into a region

Common mistakes

Before you move on

The farthest point of a half-disc

Questions