Maths › Further Pure 2 › Further loci and regions in the Argand diagram
Further loci and regions in the Argand diagram
Ratios of distances draw circles you would not expect, and a fixed angle between two directions traces an arc. Combine the conditions and a region appears.
Builds on Modulus, argument and loci and Circles.
IN THIS TOPIC
- Identify |z − a| = k|z − b| as a circle when k ≠ 1, and find its centre and radius.
- Recognise a constant argument of a quotient as an arc of a circle.
- Shade regions defined by combined inequalities.
WHAT YOU PROBABLY THINK
|z − a| = k|z − b| is always a perpendicular bisector, whatever k is.
Circles from a ratio of distances
When k = 1 the locus really is the perpendicular bisector of a and b, but that is the one exceptional case. For any other k the cartesian algebra produces an x² + y² term with a coefficient other than 1, and after dividing through and completing the square a circle appears: the circle of Apollonius. The opening claim generalises from the single case where the quadratic terms happen to cancel.
WORKED EXAMPLE
Finding the circle
Find the locus of points satisfying |z| = 2|z − 3|.
With z = x + iy: x² + y² = 4((x − 3)² + y²).
Expanding and collecting: 3x² + 3y² − 24x + 36 = 0, so x² + y² − 8x + 12 = 0.
Completing the square: (x − 4)² + y² = 4, a circle of centre (4, 0) and radius 2.
Sense check: z = 2 gives 2 = 2 × 1, and z = 6 gives 6 = 2 × 3. Both are on the circle, at its two ends along the real axis.
Arcs from a fixed angle
The condition arg((z − a)/(z − b)) = β fixes the angle subtended at z by the segment from b to a. By the inscribed angle theorem, the points doing that lie on an arc of a circle through a and b, on one side only, with a and b themselves excluded. An angle of π/2 gives a semicircle with ab as diameter.
WORKED EXAMPLE
A right angle traces a semicircle
Describe the locus arg((z − 1)/(z + 1)) = π/2.
The condition says the segment from −1 to 1 subtends a right angle at z, and the sign of the argument fixes which side.
The angle in a semicircle is a right angle, so the locus is the arc of the circle on diameter from −1 to 1, that is |z| = 1, taking the upper half.
Testing z = i: (i − 1)/(i + 1) = i, whose argument is π/2. The endpoints ±1 are excluded, since the quotient is then undefined or zero.
Regions come from inequalities. Conditions like α ≤ arg(z − z₁) ≤ β cut a wedge from the plane with its vertex at z₁, while p ≤ Re(z) ≤ q cuts a vertical strip. Sketch each boundary first, then shade the overlap.
YOUR TURN
A region from two conditions
Sketch the region satisfying both |z − 2i| ≤ 2 and 0 ≤ Re(z) ≤ 2, and describe its shape.
Show the working
The first is the closed disc of radius 2 centred at (0, 2).
The second is the vertical strip between x = 0 and x = 2, inclusive.
The overlap is the right half of that disc, bounded on the left by the diameter along x = 0 and cut nowhere on the right, since the disc reaches only to x = 2. The region is a half-disc of area 2π.
THE EXAM BIT
- For |z − a| = k|z − b|, square both sides and convert to cartesian; the circle only appears after completing the square.
- Say explicitly whether k = 1, since that single case gives a line rather than a circle.
- For an argument locus, state that it is an arc, not a full circle, and which endpoints are excluded.
- Shade regions after drawing every boundary, and mark whether each boundary is included.
CHECK YOURSELF
Describe the locus |z − 1| = |z + 3|.
Show a hint
Here the ratio k equals 1.
Show the answer
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|z − a| = k|z − b| is a circle unless k = 1, when it degenerates to the perpendicular bisector.
A constant argument of a quotient traces an arc through a and b, with those two points excluded.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Identify |z − a| = k|z − b| as a circle when k ≠ 1, and find its centre and radius.
- Recognise a constant argument of a quotient as an arc of a circle.
- Shade regions defined by combined inequalities.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.