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Roots of unity and complex roots questions
Every non-zero number has exactly n distinct nth roots, arranged as a regular polygon on the Argand diagram. Find one, and rotation finds the rest.
7 original questions · 31 marks · the roots of unity and complex roots notes · Complex numbers
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State how many solutions z⁵ = 32 has over the complex numbers, and give the modulus of each.
Worked answer
Exactly five, the fifth roots of 32. Each has modulus 321/5 = 2; B1 for five solutions, B1 for the modulus. they differ only in argument, spaced 2π/5 apart round the circle of radius 2.List the arguments of the sixth roots of unity, describe the shape they make on an Argand diagram, and state their sum.
Worked answer
Arguments 0, π/3, 2π/3, π, −2π/3, −π/3, which mark a regular hexagon on the unit circle with one vertex at 1. Their sum is 0, by the rotational symmetry of the set. B1 for the six arguments, B1 for the regular hexagon, B1 for the sum being zero.Solve z³ = −27, giving all roots exactly.
Worked answer
−27 has modulus 27 and argument π, so one root has modulus 3 and argument π/3, that is z = 3(cos π/3 + i sin π/3) = 3/2 + (3√3/2)i. The others sit 2π/3 round, at z = −3 (argument π) and the conjugate 3/2 − (3√3/2)i. M1 for the modulus and argument of −27, A1 for the first root, M1 for spacing the others 2π/3 apart, A1 A1 for the remaining roots. Three roots, an equilateral triangle with the real root at −3.Solve z⁴ = −16, giving all roots in the form a + bi with exact values, and describe their arrangement.
Worked answer
Modulus 16 and argument π, so each root has modulus 2 with arguments π/4, 3π/4, −3π/4, −π/4, giving z = ±√2 ± √2 i, all four sign combinations. They form a square of circumradius 2 with vertices on neither axis, one in each quadrant. M1 for modulus 2 and argument π/4, A1 for the four arguments, M1 for converting to a + bi, A1 for the four roots, B1 for the square.By considering the factorisation of z⁵ − 1, show that the five fifth roots of unity sum to zero.
Worked answer
z⁵ − 1 = (z − 1)(z⁴ + z³ + z² + z + 1). Any root other than 1 kills the second bracket, so the four non-trivial roots satisfy z⁴ + z³ + z² + z + 1 = 0; adding 1's contribution, the full sum of all five roots is the negated z⁴-coefficient of the quintic, which is 0. M1 for the factorisation, A1 for the quartic satisfied by the other four roots, A1 for the sum being zero.Solve z³ = 64i, giving all roots exactly.
Worked answer
64i has modulus 64 and argument π/2, so one root has modulus 4 and argument π/6, that is z = 4(cos π/6 + i sin π/6) = 2√3 + 2i. Spacing the rest 2π/3 apart gives arguments 5π/6 and −π/2, so z = −2√3 + 2i and z = −4i. M1 for the modulus and argument of 64i, A1 for the first root, M1 for spacing the others 2π/3 apart, A1 A1 for the remaining roots. Checking the last one, (−4i)³ = −64i³ = 64i.w = cos(2π/5) + i sin(2π/5). Show that 1 + w + w² + w³ + w⁴ = 0, and hence show that cos(2π/5) + cos(4π/5) = −1/2.
Worked answer
w is a fifth root of unity other than 1, since w⁵ = cos 2π + i sin 2π = 1 and w ≠ 1. Factorising, w⁵ − 1 = (w − 1)(1 + w + w² + w³ + w⁴) = 0, and w − 1 ≠ 0, so the second bracket vanishes. By de Moivre, wk = cos(2kπ/5) + i sin(2kπ/5). Taking real parts of the sum gives 1 + cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = 0. Now cos(6π/5) = cos(4π/5) and cos(8π/5) = cos(2π/5), because w³ and w⁴ are the conjugates of w² and w, so the sum is 1 + 2cos(2π/5) + 2cos(4π/5) = 0, giving cos(2π/5) + cos(4π/5) = −1/2. B1 for w⁵ = 1 with w ≠ 1, M1 for the factorisation, A1 for the bracket vanishing, M1 for de Moivre on each power, A1 for the real part, M1 for pairing the conjugate cosines, A1 for the equation in two cosines, A1 for the printed result. Pairing the roots with their conjugates is what halves the work; the imaginary parts cancel in the same pairs.
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