Questions on modulus, argument and loci

Question 1

Find the modulus and argument of z = −1 + √3 i.

Accepted Answer

|z| = √(1 + 3) = 2. The point (−1, √3) sits in the second quadrant, so arg z = π − π/3 = 2π/3. B1 for the modulus, M1 for the quadrant, A1 for the argument. A sketch first; the calculator's arctan alone would report the wrong quadrant.

Question 2

Write the number with modulus 2 and argument π/3 in the form x + yi, exactly.

Accepted Answer

x = 2 cos(π/3) = 1 and y = 2 sin(π/3) = √3, so z = 1 + √3 i. M1 for x = r cos θ and y = r sin θ, A1 for 1 + √3 i.

Question 3

Describe the locus of points z on an Argand diagram satisfying arg(z + 2) = −π/3, and state whether the point −2 belongs to it.

Accepted Answer

Write the condition as arg(z − (−2)) = −π/3: the direction from −2 to z is fixed, so the locus is a half-line starting at the point (−2, 0), making an angle of π/3 below the positive real direction. B1 for the half-line with its starting point and direction. The point −2 itself is not on the locus, because z = −2 makes z + 2 zero, and zero has no argument. B1 for the excluded end. A half-line, not a full line: the opposite direction, 2π/3, is a different locus, and the end is marked with an open circle.

Question 4

z has modulus 5 and argument π/3. Write down the modulus and the argument of z*, of 2z and of −z, giving each argument in the interval −π < θ ≤ π.

Accepted Answer

Conjugation reflects in the real axis, so z* has modulus 5 and argument −π/3. B1. Multiplying by the positive real number 2 stretches from the origin without turning, so 2z has modulus 10 and argument π/3. B1. Negating turns through half a turn, giving modulus 5 and argument π/3 + π = 4π/3, which is outside the principal interval; subtracting 2π gives −2π/3. B1.

Question 5

z has modulus 2 and argument π/6; w has modulus 3 and argument π/4. Find the modulus and argument of zw and of z/w.

Accepted Answer

Multiplying multiplies moduli and adds arguments: |zw| = 6, arg zw = π/6 + π/4 = 5π/12. Dividing divides and subtracts: |z/w| = 2/3, arg(z/w) = π/6 − π/4 = −π/12. B1 B1 for the modulus and argument of zw, B1 B1 for those of z/w.

Question 6

Sketch the locus |z − 3 − 4i| = 2, and find the greatest and least values of |z| on it.

Accepted Answer

A circle of radius 2 centred at 3 + 4i, the point (3, 4). The centre is distance 5 from the origin, so |z| runs from 5 − 2 = 3 to 5 + 2 = 7, along the line joining the origin to the centre. B1 for the centre, B1 for the radius, M1 for the distance of the centre from the origin, A1 for the two values.

Question 7

Describe the locus |z − 2| = |z + 2| and justify the description.

Accepted Answer

Equal distance from 2 and from −2 gives the perpendicular bisector of the segment joining (2, 0) and (−2, 0), which is the imaginary axis. B1 for equal distances, B1 for the perpendicular bisector, B1 for naming the imaginary axis.

Question 8

On one Argand diagram, shade the region satisfying both |z| ≤ 3 and 0 ≤ arg z ≤ π/2, and find its area.

Accepted Answer

|z| ≤ 3 is the disc of radius 3, and the argument condition keeps the first quadrant, including both bounding axes. The region is a quarter disc of area (1/4)π(3²) = 9π/4, about 7.07. B1 for the disc, B1 for the argument region, M1 for a quarter of the disc area, A1 for 9π/4.

Question 9

z = 2 cis(2π/3) and w = −√3 + i. Find the modulus and argument of w, and hence find zw, giving it in the form r cis θ with −π < θ ≤ π and in the form x + yi.

Accepted Answer

|w| = √(3 + 1) = 2, and (−√3, 1) sits in the second quadrant, so arg w = π − π/6 = 5π/6. B1 for the modulus and argument of w. Multiplying multiplies the moduli and adds the arguments: |zw| = 2 × 2 = 4 and arg zw = 2π/3 + 5π/6 = 3π/2. M1 for the product of moduli with the sum of arguments. An argument of 3π/2 is outside −π < θ ≤ π, and subtracting 2π gives −π/2, so zw = 4 cis(−π/2). A1 for the reduced argument. Converting, 4(cos(−π/2) + i sin(−π/2)) = 4(0 − i) = −4i. A1.

Question 10

Find the two points at which the circle |z| = 13 meets the locus |z| = |z − 10|, giving each in the form x + yi.

Accepted Answer

|z| = |z − 10| holds for the points equidistant from 0 and 10, which is the perpendicular bisector of the segment joining (0, 0) and (10, 0), the vertical line x = 5. B1 for the line x = 5. On the circle, x² + y² = 169, so 25 + y² = 169. M1 for substituting. Then y² = 144 and y = ±12. A1. The meeting points are 5 + 12i and 5 − 12i. A1. There are exactly two because the line x = 5 passes 5 from the centre, less than the radius 13.

Question 11

Shade on an Argand diagram the region satisfying both |z − 3i| < 3 and |z| ≥ |z − 4|, stating which boundary is drawn dashed and which solid, and state with a reason whether the point 3 + 3i lies in the region.

Accepted Answer

|z − 3i| = 3 is the circle of centre (0, 3) and radius 3, and the inequality is strict, so it is drawn dashed and the inside is wanted. B1. |z| ≥ |z − 4| holds where the distance from 0 is at least the distance from 4: squaring, x² + y² ≥ x² − 8x + 16 + y², so 8x ≥ 16 and x ≥ 2, the closed half plane to the right of the solid line x = 2. B1. The region is the overlap, a piece of the disc between x = 2 and x = 3; the origin fails x ≥ 2, so it is not in it, while 2.5 + 3i satisfies both. B1 for the shaded overlap. The point 3 + 3i gives |3 + 3i − 3i| = 3, which is not less than 3, so it lies on the dashed circle and is not in the region, even though x = 3 satisfies the other condition. B1.

Question 12

The locus of z satisfies |z − 2| = 2|z + 1|. Show that it is a circle, and find its centre and radius.

Accepted Answer

Put z = x + iy and square both sides, which is safe because both are non-negative. Then (x − 2)² + y² = 4[(x + 1)² + y²]. Expanding gives x² − 4x + 4 + y² = 4x² + 8x + 4 + 4y², so 0 = 3x² + 3y² + 12x. Dividing by 3 gives x² + y² + 4x = 0, and completing the square gives (x + 2)² + y² = 4, a circle of centre (−2, 0) and radius 2. M1 for putting z = x + iy and squaring, A1 for the expanded equation, M1 for collecting terms, A1 for x² + y² + 4x = 0, M1 for completing the square, A1 for the centre and radius. An unequal ratio of distances always produces a circle, in contrast with |z − a| = |z − b|, where the squared terms cancel and a straight line is left. The centre is not the midpoint of 2 and −1, so it cannot be written down by inspection.

Question 13

The complex number z satisfies |z − 4| = 2. Find the greatest and least values of arg z, and find the two points of the locus at which they occur, giving each in the form x + yi.

Accepted Answer

The locus is the circle of centre (4, 0) and radius 2. The origin is 4 from the centre and 4 > 2, so the origin lies outside the circle and the rays from the origin that give the extreme arguments are the two tangents. B1 for the origin being outside, with the extremes at the tangents. A tangent meets the radius at the point of contact at a right angle, giving a right-angled triangle with hypotenuse 4, from the origin to the centre, and opposite side 2, the radius. M1 for that triangle. So sin α = 2/4 = ½, where α is the angle that tangent makes with the real axis at the origin. A1. Hence α = π/6, and by symmetry in the real axis the greatest value of arg z is π/6 and the least is −π/6. A1. The tangent length is √(4² − 2²) = 2√3, so a point of contact is 2√3 from the origin at argument ±π/6. M1 for the tangent length. That gives 2√3(cos(π/6) ± i sin(π/6)) = 3 + √3 i and 3 − √3 i. A1. Checking, |3 + √3 i − 4| = |−1 + √3 i| = 2.

Question 14

The complex number z satisfies both arg(z − 2) = π/4 and |z − 2| ≤ 3√2. Find the Cartesian equation of the set of such z, stating the range of x, and find the greatest value of |z| and the point at which it occurs.

Accepted Answer

arg(z − 2) = π/4 is the half-line from (2, 0) at π/4, with (2, 0) excluded, and |z − 2| ≤ 3√2 cuts it off 3√2 from that end. B1 for the half-line from 2 with the end excluded. A point on it is z = 2 + t(cos π/4 + i sin π/4) = 2 + t/√2 + (t/√2)i with 0 < t ≤ 3√2, so x = 2 + t/√2 and y = t/√2. M1 for a general point. Eliminating t gives y = x − 2, and t = 3√2 gives t/√2 = 3, so 2 < x ≤ 5. A1 for the equation with the range. Then |z|² = x² + (x − 2)² = 2x² − 4x + 4 = 2(x − 1)² + 2. M1. That increases for x > 1, so on 2 < x ≤ 5 it is greatest at the closed end x = 5. A1 for the greatest value being at the end of the segment. There |z|² = 34, so the greatest value is |z| = √34, about 5.83, at z = 5 + 3i. A1.

Question 15

A transmitter at the origin has a range of 10 km, and a second transmitter at the point 12 + 5i has a range of 8 km, where one unit on the Argand diagram is 1 km. A receiver at z can use a transmitter when it is within that transmitter's range. Write down the two conditions for a receiver to be able to use both, show that the set of such points is not empty and that neither transmitter's coverage lies entirely inside the other's, and determine whether a receiver at 6 + 2i can use both.

Accepted Answer

The two conditions are |z| ≤ 10 and |z − 12 − 5i| ≤ 8, each a closed disc about a transmitter. B1 for both. The distance between the transmitters is |12 + 5i| = √(144 + 25) = √169 = 13 km. M1 for the distance, A1 for 13 km. Now 10 − 8 = 2 and 10 + 8 = 18, and 2 < 13 < 18: the centres are closer together than the sum of the radii, so the discs overlap and the set of points is not empty, and further apart than the difference, so neither disc lies inside the other. B1 for both comparisons. For the receiver, |6 + 2i| = √40 = 2√10 ≈ 6.32 km, which is at most 10, and |6 + 2i − 12 − 5i| = |−6 − 3i| = √45 = 3√5 ≈ 6.71 km, which is at most 8. M1 for testing both distances. Both hold, so the receiver can use both transmitters. A1.

Question 16

The diagram shows the points P and Q, which represent the complex numbers z and w. Write down z and w, find the modulus and the argument of each, and hence find zw in the form r cis θ, with −π < θ ≤ π.

Accepted Answer

P is 2 to the right and 2 up, so z = 2 + 2i, and Q is 3 to the left and 3 up, so w = −3 + 3i. B1 for both numbers read from the diagram. |z| = √(2² + 2²) = √8 and |w| = √((−3)² + 3²) = √18. M1 for the modulus of either. So |z| = 2√2 and |w| = 3√2. A1 for both moduli. P lies in the first quadrant, so arg z = arctan(2/2) = π/4; Q lies in the second quadrant, so arg w = π − arctan(3/3) = π − π/4 = 3π/4. M1 for using the quadrant of each point to place its argument, A1 for both arguments. Multiplying multiplies the moduli and adds the arguments: |zw| = 2√2 × 3√2 = 6 × 2 = 12 and arg(zw) = π/4 + 3π/4 = π, so zw = 12 cis π. A1 for the product. Multiplying directly checks it: (2 + 2i)(−3 + 3i) = −6 + 6i − 6i + 6i² = −12, and −12 is indeed 12 cis π.