Draw argument rays and combine angular and radial conditions
Interpret a shifted argument as a ray from a fixed point, exclude its undefined origin and intersect angular conditions with radial bounds.
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 arguments, shifts and radian quadrants.
- Describe disks, annuli and intersections of regions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What parameter condition gives a ray from a in direction α?
What you will learn
- Describe a shifted-argument locus as a ray with the correct direction and excluded endpoint.
- Combine radial and angular inequalities with explicit boundary membership.
An argument specifies a ray, not an entire line
The equation arg(z − a) = α describes points whose displacement from a has direction α. Write z = a + rcis α with r > 0; the starting point a is excluded because arg 0 is undefined. For a = 1 + i and α = π/4, the ray has Cartesian equation y = x with x > 1. The part of that line with x < 1 points in the opposite direction and does not belong.
Direction matters even for a vertical ray. The condition arg(z + 1 − 2i) = −π/2 starts at a = −1 + 2i and goes downwards: x = −1 and y < 2. For arg(z − 2) = 3π/4, the ray starts at (2,0), goes up and left and has y = 2 − x with x < 2. Always state the direction restriction alongside an infinite-line equation.
Intersect a sector with the permitted distances
Consider 1 < |z − (1 + i)| ≤ 3 and 0 ≤ arg(z − (1 + i)) ≤ π/2, using the usual principal argument. The radial condition selects distances greater than one and at most three from (1,1); the angular condition selects directions from rightwards to upwards, including both rays. The intersection is a quarter-annular region to the upper right of the centre. Its inner arc is excluded, outer arc included and both radial sides included only between the permitted distances.
Test all conditions for a proposed boundary point. The point 2 + i has distance one and is excluded despite lying on an included angular ray. The point 4 + i has distance three and angle zero, so it is included. The point 1 + 3i has distance two and angle π/2, so it is included; 3 − i has a negative argument and is excluded. The centre is excluded by the radial bound and independently has no argument. If a sector spans an argument interval endpoint, split or normalise the angular description carefully rather than treating a discontinuity in principal angles as a missing geometric direction.
Pause and explain
For arg(z − (1 + i)) = π/4, which Cartesian description is correct?
Worked example
Describe arg(z − (1 + i)) = π/4 in Cartesian terms and specify its endpoint.
Show the worked solution
- The displacement starts at (1,1) and points π/4 anticlockwise from the positive real direction.
- Its components are equal and positive, giving y = x with x > 1.
- The start (1,1) is excluded because its displacement is zero and has no argument.
Answer Ray y = x, x > 1, from excluded endpoint (1,1), going up and right.
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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 arg(z + 1 − 2i) = −π/2, giving its start, Cartesian condition and excluded point.
Give me a hint
Read the shifted centre first, then take the downward imaginary direction.
Compare my reasoning
- The start is a = −1 + 2i, or (−1,2).
- A downward displacement gives x = −1 and y < 2.
- The starting point itself is excluded because arg 0 is undefined.
Downward ray x = −1, y < 2, from excluded endpoint (−1,2).
Look for these in your work
- Read both signs of the shifted starting point.
- Specify the ray direction and exclude its endpoint.
Describe arg(z − 2) = 3π/4 in Cartesian form, specifying the direction restriction.
Give me a hint
The quadrant II displacement has equal magnitude real and imaginary components with opposite signs.
Compare my reasoning
- The starting point is (2,0).
- Direction 3π/4 means x − 2 < 0 and y = −(x − 2).
- Thus the ray is y = 2 − x with x < 2, and its start is excluded.
Ray y = 2 − x, x < 2, from excluded endpoint (2,0), going up and left.
Look for these in your work
- Preserve the shift when forming the line equation.
- Retain the one-sided direction and excluded endpoint.
Describe 1 < |z − (1 + i)| ≤ 3 and 0 ≤ arg(z − (1 + i)) ≤ π/2. Decide whether 2 + i, 4 + i, 1 + 3i and 3 − i belong.
Give me a hint
Intersect the radial annulus and the two included angular rays.
Compare my reasoning
- The region is the upper-right quarter-annulus centred at (1,1), inner radius 1 excluded and outer radius 3 included; the radial sides are included over 1 < r ≤ 3.
- 2 + i has r = 1 and is excluded; 4 + i has r = 3 and angle 0, so it is included.
- 1 + 3i has r = 2 and angle π/2, so it is included; 3 − i has negative argument −π/4 and is excluded.
Quarter-annulus centred at (1,1), 1 < r ≤ 3, 0 ≤ θ ≤ π/2; 2 + i excluded, 4 + i included, 1 + 3i included, 3 − i excluded.
Look for these in your work
- Describe all radial and angular boundaries separately.
- Test each point against both conditions and exclude undefined arguments.
Common mistakes
- An argument equality defines a ray; the opposite half of the line has a different direction.
- Exclude the shifted origin wherever an argument is used.
- An angular boundary is included only where all radial conditions also hold.
What can you explain now?
Does z = a belong to arg(z − a) = α?
Compare with the explanation
No.
Its displacement is zero and the argument is undefined, for every α.
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Make it stick
Turn a shifted argument into a directed ray, exclude its origin and describe an intersection of angular and radial conditions.
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