Describe modulus circles disks and annular regions
Interpret |z − a| as distance from a fixed point, draw circular boundaries and distinguish strict and inclusive radial inequalities.
Section 3.9 · Complex numbers
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.9. Paper 1 algebra and radian trigonometry are assumed. Use Cartesian working for products, quotients and the two square roots of a non-zero complex number. Arguments usually use −π < θ ≤ π; an alternative interval 0 ≤ θ < 2π is acceptable unless specified. The argument of zero is undefined. General nth-root methods are beyond this mapped sequence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Interpret complex differences as displacements.
- Use the equation of a circle and inequality boundaries.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does |z − a| measure on an Argand diagram?
What you will learn
- Convert a fixed-distance locus to a circle and Cartesian equation.
- Describe disks and annuli with the correct included or excluded boundaries.
A shifted modulus measures distance from its centre
For a = u + iv and z = x + iy, |z − a| = √[(x − u)² + (y − v)²]. Thus |z − a| = k, for k > 0, is the circumference of a circle centred at (u,v) with radius k. The equation |z + 2 − i| = 3 uses z − (−2 + i), so its centre is (−2,1) and Cartesian equation is (x + 2)² + (y − 1)² = 9. Read the whole shifted value before naming the centre.
The locus |z − (1 + i)| < 2 is the open disk centred at (1,1), not just its circumference. Its circular boundary is excluded, while its centre is included because distance zero is less than two. Use a dashed or otherwise clearly labelled excluded boundary. The inclusive condition ≤ 2 adds the circumference; a condition > 2 selects the exterior and excludes the circumference.
Two radial bounds define an annulus
The conditions 1 ≤ |z − (2 − i)| ≤ 3 give an annulus centred at (2,−1), with both radius-one and radius-three boundaries included. Its centre is excluded by the lower bound. A strict lower bound with inclusive upper bound excludes the inner circle but includes the outer one. Describe each boundary separately rather than calling the whole region open or closed without checking both inequalities.
Test membership using squared distance where radii are non-negative: distance < √5 is equivalent to squared distance < 5. At a = −1 + 2i, point 1 + 3i has squared displacement 4 + 1 = 5, so it lies on the boundary and fails a strict inequality. The centre has distance zero and satisfies the open-disk condition. Negative distance bounds need a direct check: |z − a| < −1 has no solutions because modulus cannot be negative.
Pause and explain
For |z − (1 + i)| < 2, which boundary statement is correct?
Worked example
Describe and sketch |z − (1 + i)| < 2, including its boundary.
Show the worked solution
- The shifted modulus is distance from (1,1).
- The condition selects every point less than two units from that centre.
- Shade the interior and exclude the radius-two circumference; the centre itself is included.
Answer Open disk centred at (1,1), radius 2; circumference excluded, centre included.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe |z + 2 − i| = 3 and write its Cartesian equation for z = x + iy.
Give me a hint
Rewrite the shifted term as z minus the centre’s complex value.
Compare my reasoning
- Use z + 2 − i = z − (−2 + i).
- The centre is (−2,1) and the radius is 3.
- Squaring the distance gives (x + 2)² + (y − 1)² = 9, describing the circumference.
Circle centred at (−2,1), radius 3; (x + 2)² + (y − 1)² = 9.
Look for these in your work
- Read the centre with the correct shifted signs.
- Distinguish the circumference from its interior.
Describe 1 ≤ |z − (2 − i)| ≤ 3, specifying the centre and both boundaries.
Give me a hint
Apply each distance bound separately.
Compare my reasoning
- The centre is (2,−1).
- Permitted distances are from 1 to 3 inclusive, so the region is an annulus.
- Both circular boundaries are included, while the central disk of radius less than 1 is excluded.
Annulus centred at (2,−1), inner radius 1 and outer radius 3; both boundaries included.
Look for these in your work
- Identify both radii and the signed centre.
- Explain membership at each boundary and at the centre.
For |z − (−1 + 2i)| < √5, decide whether z = 1 + 3i, z = −1 + 2i and z = 2 + 2i belong. Explain each decision.
Give me a hint
Compare each squared displacement with 5; equality fails a strict inequality.
Compare my reasoning
- At 1 + 3i the displacement is 2 + i with squared length 5, so this boundary point is excluded.
- At −1 + 2i the displacement is zero, so the centre is included.
- At 2 + 2i the displacement is 3 with squared length 9 > 5, so it is outside and excluded.
1 + 3i excluded (boundary); −1 + 2i included (centre); 2 + 2i excluded (outside).
Look for these in your work
- Calculate distances from the supplied centre.
- Respect the strict boundary rather than accepting equality.
Common mistakes
- Read z − a carefully to avoid reversing the centre’s signs.
- An inequality describes a region, while equality describes its boundary.
- Test both annular boundaries and do not infer a direction from modulus alone.
What can you explain now?
Does |z − a| < −1 have any solutions?
Compare with the explanation
No.
A distance is non-negative, so it cannot be less than a negative bound.
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Make it stick
Translate a shifted modulus equality or inequality into a circle, disk or annulus, stating each boundary’s membership.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.