Questions on further loci and regions in the Argand diagram

Question 1

Write down the centre and radius of the circle |z + 2 − i| = 3, and find the cartesian equation of the locus |z − 4| = |z − 2i|.

Accepted Answer

Rewrite the first as |z − (−2 + i)| = 3, so the centre is (−2, 1) and the radius is 3. The sign trap is real: |z + 2 − i| centres at −2 + i, not at 2 − i.
The second says z is equidistant from 4 and from 2i, so the locus is the perpendicular bisector of the segment joining (4, 0) and (0, 2). That segment has gradient −1/2 and midpoint (2, 1), so the bisector has gradient 2 and equation y = 2x − 3. B1 B1 for the centre and radius, B1 for the equation of the bisector.

Question 2

Sketch and describe the locus arg(z − 2) = π/4.

Accepted Answer

It is a half-line starting at the point 2 on the real axis and running up and to the right at 45° to the positive real direction, that is the part of y = x − 2 with x > 2. The starting point itself is excluded, because the argument of zero is undefined. B1 for the half-line, B1 for the start at 2, B1 for the direction. A condition on arg(z − a) alone always gives a half-line, never the whole line.

Question 3

Write down the centre and radius of the circle |z − 3 + 4i| = 5, and state whether the origin lies on it.

Accepted Answer

Rewrite the equation as |z − (3 − 4i)| = 5, so the centre is 3 − 4i, the point (3, −4), and the radius is 5. B1 for the centre and radius. The distance from the origin to (3, −4) is √(9 + 16) = 5, which is exactly the radius, so the origin does lie on the circle. B1 for that test. Read the centre off the bracket as it stands: the sign of every term flips, so +4i in the equation puts the centre at −4i.

Question 4

Shade on an Argand diagram the region defined by |z + 2| ≤ 2, and state whether the origin belongs to the region.

Accepted Answer

The condition is |z − (−2)| ≤ 2, so the region is the closed disc of centre (−2, 0) and radius 2, shaded inside the circle. B1 for the centre and radius, B1 for shading the inside with the boundary drawn as a solid curve. The origin is at distance 2 from the centre, so it lies on the boundary, and the inequality is a weak one, so the origin does belong. B1 for the origin with a reason. Which boundary points belong is part of the answer, not decoration: a strict inequality would exclude the whole circle and is drawn as a broken curve.

Question 5

Describe the locus given by arg((z − 1)/(z + 1)) = π/2, justifying the shape.

Accepted Answer

The argument of a quotient is the difference of the arguments, so the condition says the vector from 1 to z turns through a right angle from the vector from −1 to z. The angle subtended at z by the segment joining −1 and 1 is therefore 90°, and the angle in a semicircle puts z on the circle with that segment as diameter: |z| = 1. The sign of the argument selects the upper semicircle, and the two endpoints are excluded. M1 for the difference of arguments, A1 for the right angle at z, M1 for the angle in a semicircle, A1 for the upper semicircle with ends excluded.

Question 6

Shade the region satisfying both |z| ≤ 4 and |z − 3| ≥ |z + 3|, and find its area.

Accepted Answer

The first condition is the disc of radius 4 centred at the origin, boundary included. The second says z is at least as far from 3 as from −3, which is the half-plane x ≤ 0. The region is the left half of the disc, a semicircular area of 8π, about 25.13 square units. B1 for the disc, M1 A1 for the half-plane x ≤ 0, B1 for the area.

Question 7

Find the cartesian equation of the locus |z − 3| = 2|z + 3| and describe the locus geometrically.

Accepted Answer

Write z = x + iy and square both sides, which clears the roots in one step: (x − 3)² + y² = 4[(x + 3)² + y²]. Square the modulus before expanding; squaring term by term is not valid.
Expanding and collecting gives 3x² + 30x + 27 + 3y² = 0, so x² + 10x + 9 + y² = 0.
Completing the square gives (x + 5)² + y² = 16, a circle of centre (−5, 0) and radius 4. M1 for squaring both sides, A1 for the expanded equation, M1 for completing the square, A1 for the circle equation, A1 for the centre and radius. A ratio of distances gives a circle whenever the ratio is not 1; only the ratio 1 produces a straight line.

Question 8

The region R is defined by |z + 4| ≤ 3|z − 4|. Find the cartesian equation of the boundary of R, and determine, by testing a suitable point, whether R is the inside or the outside of that boundary.

Accepted Answer

Square both sides of the boundary equation |z + 4| = 3|z − 4| with z = x + iy: (x + 4)² + y² = 9[(x − 4)² + y²]. Expanding and collecting gives 8x² + 8y² − 80x + 128 = 0, so x² + y² − 10x + 16 = 0, and completing the square gives (x − 5)² + y² = 9, a circle of centre (5, 0) and radius 3.
Now test a point. At z = 0 the inequality reads 4 ≤ 12, which is true, and 0 is at distance 5 from the centre, outside the circle. So R is the outside of the circle, boundary included since the inequality is weak. The instinct that ≤ must mean the inside is exactly what the test point corrects: the point 4 sits inside the circle and fails the condition, because 8 ≤ 0 is false. M1 for squaring both sides, A1 for the collected equation, A1 for the circle with its centre and radius, M1 for testing a point, A1 for the exterior with the boundary included.

Question 9

Shade on an Argand diagram the region satisfying both |z − 2 − 2i| ≤ 2 and Re(z) ≥ 2, and find the exact area of that region.

Accepted Answer

The first condition is the closed disc of centre (2, 2) and radius 2. The second is the closed half-plane to the right of the vertical line x = 2. B1 for the disc, B1 for the half-plane.
That line runs through the centre of the disc, so it cuts the disc into two halves and the region is the right-hand half. M1 for recognising that the line is a diameter. The area is half of π × 2², that is 2π, about 6.283 square units. A1 for 2π. Both boundaries belong to the region, since both inequalities are weak, so the semicircular arc and the straight edge are drawn solid.

Question 10

The point z moves on the locus |z − 6| = |z − 2i|. Find the cartesian equation of the locus, and hence find the least value of |z| and the complex number at which it occurs, giving exact answers.

Accepted Answer

The condition says z is equidistant from 6 and from 2i, so the locus is the perpendicular bisector of the segment joining (6, 0) and (0, 2). That segment has midpoint (3, 1) and gradient −1/3, so the bisector has gradient 3 and passes through (3, 1): its equation is y = 3x − 8. M1 for the perpendicular bisector, A1 for the equation.
The least value of |z| is the shortest distance from the origin to that line, which is measured along the perpendicular from the origin. Writing the line as 3x − y − 8 = 0, the distance is 8/√(3² + 1²) = 8/√10 = 4√10/5, about 2.530. M1 for the perpendicular distance.
The foot of that perpendicular lies on the line through the origin with gradient −1/3, that is y = −x/3. Solving with y = 3x − 8 gives 3x − 8 = −x/3, so 10x = 24 and x = 12/5, y = −4/5. The least modulus occurs at z = (12 − 4i)/5, that is 2.4 − 0.8i, and |2.4 − 0.8i| = √(5.76 + 0.64) = √6.4 = 4√10/5 agrees. A1 for the point.
Sketch the line before anything else: it crosses the real axis at 8/3 and the imaginary axis at −8, so the nearest point sits below the real axis, and a candidate with positive imaginary part is the sign of a slip. The extremes of |z| on a line come from a perpendicular, not from an intercept, and there is no greatest value, since the line runs off without end.

Question 11

Show that the locus |z − 1| = |z − i| passes through the origin, and find the two points at which it meets the circle |z| = 2, giving exact answers.

Accepted Answer

The condition says z is the same distance from 1 as from i, so the locus is the perpendicular bisector of the segment joining (1, 0) and (0, 1). That segment has midpoint (½, ½) and gradient −1, so the bisector has gradient 1 and passes through (½, ½): its equation is y = x. At the origin the two distances are |0 − 1| = 1 and |0 − i| = 1, so the origin lies on it. M1 for the perpendicular bisector, A1 for y = x, B1 for the origin.
Substituting y = x into x² + y² = 4 gives 2x² = 4, so x = ±√2 and the meeting points are √2 + √2 i and −√2 − √2 i. A1 for both points.

Question 12

The point z moves on the circle |z − 4| = 2. Find the greatest and least values of |z|, and the greatest and least values of arg z, giving exact answers.

Accepted Answer

The circle has centre 4 and radius 2, and the origin lies outside it since 4 > 2. Sketch it before anything else; the extremes are read off the diagram, not conjured from algebra.
For the modulus, the nearest and furthest points lie on the line through the origin and the centre, so the values are 4 − 2 = 2 and 4 + 2 = 6.
For the argument, the extreme positions are where a line through the origin touches the circle. The radius to the point of contact is perpendicular to that tangent, giving a right-angled triangle with hypotenuse 4, the distance to the centre, and opposite side 2, the radius. So sin θ = 2/4 = 1/2 and θ = π/6.
By symmetry in the real axis the greatest argument is π/6 and the least is −π/6. The tangent length itself is √(4² − 2²) = 2√3, which is worth quoting as evidence that the right-angled triangle was used.
Adding or subtracting the radius, as for the modulus, does not give the extreme arguments. B1 for the sketch, M1 A1 for the two moduli, M1 for the right-angled triangle, A1 for sin θ = 1/2, A1 for the two arguments.

Question 13

Describe the locus arg((z − 4)/z) = π/4, giving the centre and radius of the circle on which it lies, and stating exactly which points of that circle belong to the locus.

Accepted Answer

The condition fixes the angle that the segment from 0 to 4 subtends at z: arg(z − 4) − arg z = π/4. By the inscribed angle theorem the points doing that lie on an arc of a circle through 0 and 4.
The angle at the centre is twice the angle at the circumference, so the radii to 0 and to 4 meet at 2 × π/4 = π/2, and the triangle they form with the chord is right-angled and isosceles: R² + R² = 4², giving R = 2√2. The centre lies on the perpendicular bisector x = 2 at height √(8 − 4) = 2, so the centre is (2, 2).
Only the major arc above the real axis qualifies: at the top of the circle, z = 2 + (2 + 2√2)i, the argument difference is exactly π/4, while on the minor arc below the axis it is −3π/4, not π/4. The endpoints 0 and 4 are excluded, since the quotient is there zero or undefined. M1 for reading the condition as a subtended angle, A1 for the radius 2√2, A1 for the centre (2, 2), B1 for selecting the major arc above the axis, B1 for excluding 0 and 4.

Question 14

Shade the region satisfying both |z − 2| ≤ 2 and 0 ≤ arg z ≤ π/4, and show that its area is π + 2.

Accepted Answer

The first condition is the closed disc of centre (2, 0) and radius 2, which passes through the origin. The second is the wedge with vertex at the origin between the positive real axis and the half-line at π/4; the vertex itself is excluded because arg 0 does not exist, though a single point carries no area.
The ray at π/4 meets the circle again at 2 + 2i, since |2 + 2i − 2| = 2 and arg(2 + 2i) = π/4. The region is therefore bounded by the segment from O to (2, 2), the arc from (2, 2) round to (4, 0), and the real axis back to O.
Split it at the centre C(2, 0). The triangle O, C, (2, 2) has area ½ × 2 × 2 = 2. The radii from C to (2, 2) and to (4, 0) are perpendicular, so the remaining piece is a quarter sector of area ½ r²θ = ½ × 4 × π/2 = π. The total is π + 2, about 5.14 square units. B1 for the disc through the origin, B1 for the wedge with its vertex excluded, M1 for the intersection point 2 + 2i, M1 for splitting into triangle and sector, A1 for the areas 2 and π, A1 for π + 2.

Question 15

a and b are distinct real numbers and k is a positive constant with k ≠ 1. Show that the locus |z − a| = k|z − b| is a circle, and find its centre and radius in terms of a, b and k. Verify the result for the locus |z − 3| = 2|z + 3|.

Accepted Answer

Write z = x + iy and square both sides: (x − a)² + y² = k²[(x − b)² + y²]. M1 for squaring. Collecting terms, (1 − k²)(x² + y²) − 2(a − k²b)x + (a² − k²b²) = 0, and since k ≠ 1 the factor 1 − k² is not zero and may be divided out. A1 for the collected equation. The result carries x² and y² with equal coefficients and no xy term, which is the equation of a circle. B1 for that conclusion.
Completing the square, the centre is at ((a − k²b)/(1 − k²), 0) on the real axis, and the radius squared is (a − k²b)²/(1 − k²)² − (a² − k²b²)/(1 − k²). Putting that over the common denominator, the numerator is a²k² − 2abk² + b²k² = k²(a − b)², so the radius is k|a − b|/|1 − k²|. M1 for completing the square, A1 for the centre, A1 for the radius.
For |z − 3| = 2|z + 3|, a = 3, b = −3 and k = 2, so the centre is (3 − 4(−3))/(1 − 4) = 15/(−3) = −5 and the radius is 2 × 6/3 = 4, matching the circle |z + 5| = 4. The condition k ≠ 1 is what the whole argument rests on: at k = 1 the x² and y² terms cancel and the locus is a straight line, the perpendicular bisector of the two points.