Practise › Questions › Modulus, argument and loci
Modulus, argument and loci questions
Length and direction take over from across and up. Every complex number is a distance from the origin at an angle, multiplication scales and rotates, and equations in z draw circles and lines.
7 original questions · 26 marks · the modulus, argument and loci notes · Complex numbers
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Find the modulus and argument of z = −1 + √3 i.
Worked 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.Write the number with modulus 2 and argument π/3 in the form x + yi, exactly.
Worked 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. Modulus-argument to cartesian is two right-triangle reads.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.
Worked 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. No cartesian expansion is needed anywhere.Sketch the locus |z − 3 − 4i| = 2, and find the greatest and least values of |z| on it.
Worked 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. Centre distance plus and minus radius, once the sketch is drawn.Describe the locus |z − 2| = |z + 2| and justify the description.
Worked 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. Any equation |z − a| = |z − b| is the perpendicular bisector of a and b.On one Argand diagram, shade the region satisfying both |z| ≤ 3 and 0 ≤ arg z ≤ π/2, and find its area.
Worked 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. Sketch both loci and read the overlap; the algebra adds nothing here.The locus of z satisfies |z − 2| = 2|z + 1|. Show that it is a circle, and find its centre and radius.
Worked 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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise modulus, argument and loci one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.