Questions on roots of unity and complex roots

Question 1

State how many solutions z⁵ = 32 has over the complex numbers, and give the modulus of each.

Accepted Answer

Exactly five, the fifth roots of 32. Each has modulus 32^1/5 = 2, and they differ only in argument, spaced 2π/5 apart round the circle of radius 2. B1 for five solutions, B1 for the modulus.

Question 2

List the arguments of the sixth roots of unity, describe the shape they make on an Argand diagram, and state their sum.

Accepted 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.

Question 3

ω = e^2πi/7. Write down the seven seventh roots of unity in terms of ω, and write down the values of ω⁷ and ω¹⁰.

Accepted Answer

The roots are the powers of ω: 1, ω, ω², ω³, ω⁴, ω⁵, ω⁶. B1 for the list. Each has argument 2πk/7, so the seven are distinct, and the powers then repeat: ω⁷ = 1, and ω¹⁰ = ω⁷ω³ = ω³. B1 for both values. Powers of ω repeat with period 7, so an exponent is reduced by taking its remainder on division by 7.

Question 4

Explain why z⁴ = 0 has exactly one solution, while z⁴ = w has four distinct solutions for every non-zero complex number w.

Accepted Answer

Taking moduli, |z|⁴ = |w|. With w = 0 that gives |z| = 0, and the only number of modulus 0 is 0 itself, so z = 0 is the only solution, a root of multiplicity four rather than four distinct roots. B1. Zero also has no argument, so there is nothing to divide by 4 and no polygon to draw. B1. For non-zero w the modulus of each root is the positive real fourth root of |w|, and the arguments are (arg w)/4 increased by 0, π/2, π and 3π/2, four directions that differ by less than a whole turn, so the four numbers are distinct. B1.

Question 5

Solve z³ = −27, giving all roots exactly.

Accepted 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.

Question 6

Solve z⁴ = −16, giving all roots in the form a + bi with exact values, and describe their arrangement.

Accepted 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.

Question 7

By considering the factorisation of z⁵ − 1, show that the five fifth roots of unity sum to zero.

Accepted 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.

Question 8

Solve z³ = 64i, giving all roots exactly.

Accepted 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.

Question 9

Solve z⁴ = 8 + 8√3 i, giving the four roots in the form re^iθ with −π < θ ≤ π.

Accepted Answer

|8 + 8√3 i| = √(64 + 192) = √256 = 16, and the point (8, 8√3) is in the first quadrant with 8√3/8 = √3, so the argument is π/3. M1 for the modulus and argument, A1 for 16 and π/3. Each root has modulus 16^1/4 = 2 and one has argument (π/3)/4 = π/12, giving 2e^iπ/12. A1. The other three sit 2π/4 = π/2 further round each time, at π/12 + π/2 = 7π/12, then 13π/12, then 19π/12. M1 for spacing by π/2. The last two are outside the principal interval, and subtracting 2π gives −11π/12 and −5π/12, so the roots are 2e^7iπ/12, 2e^−11iπ/12 and 2e^−5iπ/12. A1.

Question 10

The five vertices of a regular pentagon lie on the circle |z| = 3 on an Argand diagram, and one vertex is at 3i. Find the arguments of the other four vertices, each in the interval −π < θ ≤ π, and find the equation of the form z⁵ = w that all five vertices satisfy.

Accepted Answer

The vertices are equally spaced round the circle, so consecutive arguments differ by 2π/5. M1 for the spacing. Starting from arg 3i = π/2 and adding 2π/5 each time gives 9π/10, 13π/10, 17π/10 and 21π/10. A1 for the four arguments before reduction. The last three are outside the interval, and subtracting 2π gives 9π/10, −7π/10, −3π/10 and π/10. A1. Every vertex has modulus 3, and its fifth power therefore has modulus 3⁵ = 243; multiplying an argument by 5 adds a whole number of turns to 5 × π/2, so all five fifth powers are equal. M1 for raising a vertex to the fifth power. Taking the vertex 3i, (3i)⁵ = 243i⁵ = 243i, so the equation is z⁵ = 243i. A1.

Question 11

Solve z⁶ − 7z³ − 8 = 0, giving all six roots exactly.

Accepted Answer

The equation is a quadratic in z³: (z³)² − 7z³ − 8 = 0, which factorises as (z³ − 8)(z³ + 1) = 0. M1 for treating it as a quadratic in z³. So z³ = 8 or z³ = −1. A1. For z³ = 8: modulus 8 and argument 0, so the roots have modulus 2 and arguments 0, 2π/3 and −2π/3. M1 for the cube roots of 8. Those are 2, 2(−1/2 + (√3/2)i) = −1 + √3 i and −1 − √3 i. A1. For z³ = −1: modulus 1 and argument π, so the roots have modulus 1 and arguments π/3, π and −π/3, giving 1/2 + (√3/2)i, −1 and 1/2 − (√3/2)i. A1. Six roots in all, as a sixth-degree equation must have, and they lie on two circles rather than one, because the two values of z³ have different moduli.

Question 12

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.

Accepted 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, w^k = 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.

Question 13

The five fifth roots of unity 1, ω, ω², ω³, ω⁴, where ω = e^2πi/5, are the vertices of a regular pentagon on the unit circle, and the point P represents a complex number z with |z| = 1. You may use the result that the five fifth roots of unity sum to zero. Show that the sum of the squares of the distances from P to the five vertices is 10, wherever P lies on the circle.

Accepted Answer

The distance from P to the vertex ω^k is |z − ω^k|, and for any complex number u, |u|² = uu*, so |z − ω^k|² = (z − ω^k)(z* − ω^−k), using that ω^k has modulus 1, so its conjugate is ω^−k. M1 for the squared distance as a product with the conjugate. Expanding, |z|² + 1 − zω^−k − z*ω^k. A1. Now add the five of them, k = 0 to 4. M1 for the sum over the roots. The first two terms give 5|z|² + 5, and the last two give −z(1 + ω⁻¹ + … + ω⁻⁴) − z*(1 + ω + … + ω⁴). Each bracket is the sum of all five fifth roots of unity, listed in a different order in the first case, so each is zero. A1 for both cross terms vanishing. That leaves 5|z|² + 5, and |z| = 1 gives 5 + 5 = 10. A1. Nothing in the working used the argument of z, so the total is the same for every position of P on the circle. B1.

Question 14

w is a non-zero complex number and n is an integer with n ≥ 3. Show that the n solutions of zⁿ = w are the vertices of a regular n-gon, and show that its perimeter is 2n|w|^1/n sin(π/n).

Accepted Answer

Taking moduli, |z|ⁿ = |w|, so every root has the same modulus r = |w|^1/n, the positive real nth root, and all n roots lie on one circle of that radius about the origin. B1 for the common modulus. Their arguments are (arg w)/n increased by 2πk/n for k = 0, 1, …, n − 1, so consecutive roots differ in argument by 2π/n. B1 for the equal spacing, which makes the n points a regular n-gon: equal radii and equal angles at the centre. Two consecutive roots and the origin form an isosceles triangle with two sides r and included angle 2π/n. M1 for that triangle. Dropping the perpendicular from the origin to the side bisects both the angle and the side, giving a right-angled triangle with hypotenuse r and angle π/n, so half the side is r sin(π/n) and the side is 2r sin(π/n). A1. All n sides are equal, so the perimeter is n times that. M1. Putting r = |w|^1/n gives the perimeter as 2n|w|^1/n sin(π/n). A1. The argument of w never appears in the answer, because it only turns the whole polygon; the shape and size depend on |w| and n alone.

Question 15

Show that the equation (z + 1)⁵ = z⁵ has exactly four solutions, and show that all four lie on the line Re(z) = −1/2 in the Argand diagram.

Accepted Answer

z = 0 is not a solution, since it would make the equation 1 = 0, so every solution may be divided by z⁵, giving ((z + 1)/z)⁵ = 1. M1 for dividing by z⁵. So (z + 1)/z is a fifth root of unity: (z + 1)/z = ω^k with ω = e^2πi/5 and k = 0, 1, 2, 3, 4. A1. The value k = 0 gives z + 1 = z, which no number satisfies, and each of the other four gives z(ω^k − 1) = 1 with ω^k ≠ 1, so z = 1/(ω^k − 1), exactly one number in each case. M1 for rejecting k = 0 and solving for z. Distinct powers ω, ω², ω³, ω⁴ give distinct values of z, so the equation has exactly four solutions. A1. Taking the modulus of the original equation, |z + 1|⁵ = |z|⁵, and both sides are non-negative real numbers, so |z + 1| = |z|. M1. That says the solution is the same distance from −1 as from 0, so it lies on the perpendicular bisector of the segment joining them, the vertical line Re(z) = −1/2. A1. The count agrees with the degree: the z⁵ terms cancel on expanding, leaving the quartic 5z⁴ + 10z³ + 10z² + 5z + 1 = 0. The four roots are −1/2 − (i/2)cot(kπ/5) for k = 1, 2, 3, 4, two conjugate pairs on that line.