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.
- Mark clearly which boundaries and endpoints belong to a locus.
COMMON MISCONCEPTION
|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, and that is the one exceptional case. For any other k the cartesian algebra produces an x² + y² term with a coefficient other than 1. Divide through, complete the square, and a circle appears, the circle of Apollonius. The perpendicular bisector is 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 gives (x − 4)² + y² = 4, a circle of centre (4, 0) and radius 2.
Sense check with the real points. At z = 2, 2 = 2 × 1. At z = 6, 6 = 2 × 3. Both sit 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 that the segment from b to a subtends at z. 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. Take β = π/2 and the arc is a semicircle on 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 settles which side.
The angle in a semicircle is a right angle, so the locus is an arc of the circle on the diameter from −1 to 1, that is |z| = 1, taking the upper half.
Test 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, and which boundaries count
Inequalities cut regions out of the plane. A condition α ≤ arg(z − z₁) ≤ β carves a wedge with its vertex at z₁, while p ≤ Re(z) ≤ q gives a vertical strip and |z − a| ≤ r gives a closed disc. Swapping = for ≤ in the Apollonius condition keeps the inside of that circle. Sketch every boundary first, then shade the overlap.
Marks go on the boundaries as often as on the shading. A weak inequality means a solid line and a strict one means a dashed line, and an argument condition always excludes its vertex, since arg 0 does not exist. Say which convention you are using on the diagram.
GUIDED PRACTICE
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 only reaches x = 2. The region is a half-disc of area 2π.
ASSESSMENT FOCUS
- 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 instead of a circle.
- For an argument locus, state that it is an arc and not a full circle, and name the excluded endpoints.
- Shade regions only after drawing every boundary, and mark whether each boundary is included.
- Give the centre and radius of any circle you find; naming the shape alone is not a complete answer.
CHECK YOURSELF
Describe the locus |z − 1| = |z + 3|.
Show a hint
Here the ratio k equals 1.
Show the answer
Equal distances from 1 and from −3, so this is the perpendicular bisector of the segment joining them, the vertical line x = −1.
|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.
Sketch every boundary before shading, and record which ones the region includes.
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.
Or read them with their worked answers on the further loci and regions in the argand diagram questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- 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.
- Mark clearly which boundaries and endpoints belong to a locus.
Open the full revision checklist to see every objective in the course in one place.