Loci & regions in the complex plane
Complex conditions describe familiar geometry once you read the modulus as a distance. |z − a| = r is the circle of radius r centred at a; |z − a| = |z − b| is the perpendicular bisector of the segment joining them; and arg(z − a) = θ is a RAY from a, not a full line — the half-line is the detail most often missed.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Read |z − a| as distance from aEvery modulus condition is a statement about distances, and translating it that way makes the shape obvious.
- Equal distances give a bisector|z − a| = |z − b| is the perpendicular bisector of the segment ab.
- Argument conditions give raysarg(z − a) = θ is the half-line from a (excluded) in direction θ.
- Convert to Cartesian if neededSubstitute z = x + iy and square modulus equations to get the equation of the circle or line.
Worked example
Describe the locus |z − 2i| = 3 and sketch the region |z − 1| ≤ |z + 1|.
- |z − 2i| = 3 is the set of points at distance 3 from 2i.
- That is a circle of radius 3 centred at (0, 2).
- For the second: |z − 1| ≤ |z + 1| means z is at least as close to 1 as to −1.
- The boundary |z−1| = |z+1| is the perpendicular bisector of the segment from −1 to 1, which is the imaginary axis.
Answer. A circle of radius 3 centred at 2i; and the half-plane x ≤ 0, i.e. everything on or left of the imaginary axis.
Where marks get dropped
These are the specific errors that cost credit on loci & regions in the complex plane questions — QED's rubric penalises each of them separately.
- Drawing arg(z − a) = θ as a full line. It is a ray starting at a, and the point a itself is excluded since arg 0 is undefined.
- Getting the centre’s sign wrong. |z + 3| = 2 is centred at −3, since z + 3 = z − (−3).
- Choosing the wrong side of an inequality. Test a specific point — the origin is usually easiest — to decide which region is meant.
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Loci & regions in the complex plane — frequently asked questions
What does |z − a| = k|z − b| give for k ≠ 1?
A circle, called an Apollonius circle. Only k = 1 degenerates to the perpendicular bisector line.
How do I find the Cartesian equation?
Substitute z = x + iy, take moduli as square roots of sums of squares, and square both sides. The cross terms cancel neatly for bisectors.
What region does arg z lie between two values describe?
A sector — the wedge between two rays from the origin. If the origin is excluded, say so explicitly.
The rest of Complex Numbers
Arithmetic, the Argand plane, polar form and De Moivre. Each subtopic below has its own method, worked example and mark-losing traps.
- 1The imaginary unit i & complex arithmetic
- 2Conjugates & division
- 3Modulus, argument & the Argand diagram
- 4Polar & exponential form re^{iθ}
- 5Multiplying & dividing in polar form
- 6De Moivre’s theorem & powers
- 7nth roots & roots of unity
- 8Solving polynomial equations over ℂ
- 9The fundamental theorem of algebra & conjugate roots
- 10Euler’s formula & trigonometric identities
- 11Loci & regions in the complex plane
- 12Complex multiplication as rotation & scaling
- 13Applications: phasors & AC circuits
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