Questions on De Moivre's theorem and trigonometric identities

Question 1

State De Moivre's theorem, and explain why a number must be in modulus-argument form before the theorem can be applied.

Accepted Answer

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ; for a general number, raise the modulus to the power n and multiply the argument by n. The theorem speaks about moduli and arguments, so a number given as x + yi must be converted first. B1 for stating the theorem, B1 for the reason.

Question 2

Use De Moivre's theorem to evaluate (1 − i)⁶.

Accepted Answer

1 − i has modulus √2 and argument −π/4. The sixth power has modulus (√2)⁶ = 8 and argument −3π/2, which reduces to π/2. So (1 − i)⁶ = 8i. B1 for the modulus and argument, M1 for applying De Moivre, A1 for 8i. Bracket-expansion agrees: (1 − i)² = −2i, and (−2i)³ = 8i.

Question 3

Explain why e^iθ has modulus 1 for every real θ, and write down the modulus and the argument of 3e^2iπ/3.

Accepted Answer

e^iθ = cos θ + i sin θ, so its modulus is √(cos²θ + sin²θ) = 1 for every real θ, by the Pythagorean identity. B1 for the reason. In r e^iθ the r is the modulus and the θ the argument, so 3e^2iπ/3 has modulus 3 and argument 2π/3. B1 for both.

Question 4

z = 2e^iπ/5. Find |z⁷| and arg(z⁷), giving the argument in the interval −π < θ ≤ π.

Accepted Answer

Raising to the seventh power raises the modulus to the seventh power: |z⁷| = 2⁷ = 128. B1. It multiplies the argument by 7: 7 × π/5 = 7π/5. M1. That is more than π, so subtract a whole turn: 7π/5 − 2π = −3π/5. A1.

Question 5

Evaluate (√3 + i)⁵, giving the answer in the form a + bi with exact values.

Accepted Answer

√3 + i has modulus 2 and argument π/6. The fifth power has modulus 2⁵ = 32 and argument 5π/6. Converting back: 32(cos 5π/6 + i sin 5π/6) = 32(−√3/2 + i/2) = −16√3 + 16i. B1 for the modulus and argument, M1 for the fifth power, M1 for converting back to a + bi form, A1 for −16√3 + 16i. Give exact surds: a decimal such as −27.7 + 16i is not the form asked for.

Question 6

Write down the values of (cos π/12 + i sin π/12)⁶ and (cos π/12 + i sin π/12)⁻⁶.

Accepted Answer

The power multiplies the argument and leaves the unit modulus alone, so the first is cos π/2 + i sin π/2 = i. De Moivre holds for negative n too, so the second has argument −π/2 and equals −i. B1 B1.

Question 7

By expanding (cos θ + i sin θ)³ and equating imaginary parts, show that sin 3θ = 3 sin θ − 4 sin³θ.

Accepted Answer

The binomial expansion gives cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ, and De Moivre says the total is cos 3θ + i sin 3θ. Imaginary parts: sin 3θ = 3 cos²θ sin θ − sin³θ. Replace cos²θ with 1 − sin²θ to reach sin 3θ = 3 sin θ − 4 sin³θ. M1 for the binomial expansion, A1 for it correct, M1 for equating imaginary parts, A1 for sin 3θ = 3 cos²θ sin θ − sin³θ, A1 for the printed result. The third term carries i² = −1, so 3 cos θ sin²θ belongs in the real part with a minus sign, not in the imaginary part.

Question 8

Write z = 1 + i√3 in exponential form, and hence calculate z⁴, giving the answer both as re^iθ with −π < θ ≤ π and in the form a + bi with exact values.

Accepted Answer

The modulus is √(1 + 3) = 2 and the argument is π/3, so z = 2e^iπ/3. B1 for the exponential form. Raising to the fourth power is an index law in this notation: z⁴ = 2⁴e^4iπ/3. M1 for powering both parts. An argument of 4π/3 lies outside the principal range, and subtracting a whole turn gives 16e^−2iπ/3. A1 for the reduced argument. Converting back, 16(cos(−2π/3) + i sin(−2π/3)) = 16(−1/2 − i√3/2) = −8 − 8√3i. A1 for the exact a + bi form. Leaving the argument as 4π/3 does not meet a demand for the principal range; the number is the same, the form is not the one asked for.

Question 9

θ is real with cos θ ≠ 0, and n is a positive integer. Show that 1 + i tan θ = (cos θ + i sin θ)/cos θ, and hence show that ((1 + i tan θ)/(1 − i tan θ))ⁿ = cos 2nθ + i sin 2nθ.

Accepted Answer

Writing tan θ as sin θ/cos θ and taking the factor 1/cos θ out, 1 + i tan θ = 1 + i sin θ/cos θ = (cos θ + i sin θ)/cos θ, which needs cos θ ≠ 0 both for tan θ to exist and for the division. M1 for tan θ as sin θ/cos θ with the factor taken out, A1 for the printed form. The same step gives 1 − i tan θ = (cos θ − i sin θ)/cos θ, so in the quotient the factors of 1/cos θ cancel and (1 + i tan θ)/(1 − i tan θ) = (cos θ + i sin θ)/(cos θ − i sin θ). M1 for the quotient with 1/cos θ cancelling. Now cos θ − i sin θ is the conjugate of cos θ + i sin θ, a number of modulus 1, so it is also its reciprocal: cos θ − i sin θ = (cos θ + i sin θ)⁻¹. Dividing by it therefore multiplies, and the quotient is (cos θ + i sin θ)² = cos 2θ + i sin 2θ by De Moivre. A1. Raising that to the nth power multiplies its argument by n and leaves its modulus 1 alone. M1 for applying De Moivre a second time. So ((1 + i tan θ)/(1 − i tan θ))ⁿ = (cos 2θ + i sin 2θ)ⁿ = cos 2nθ + i sin 2nθ. A1. Nothing here needed the sign of cos θ, because the awkward factor cancels before any modulus is taken; θ = π/4 checks it, where the left side is (1 + i)/(1 − i) = i and the right side is cos(π/2) + i sin(π/2) = i.

Question 10

Use e^iθe^iφ = e^i(θ + φ) to derive the formulae for cos(θ + φ) and sin(θ + φ).

Accepted Answer

Write each side in the form cos + i sin. The right side is cos(θ + φ) + i sin(θ + φ). The left side is (cos θ + i sin θ)(cos φ + i sin φ). M1 for writing both sides out. Expanding the product and replacing i² by −1 gives (cos θ cos φ − sin θ sin φ) + i(sin θ cos φ + cos θ sin φ). A1. Two complex numbers are equal exactly when their real parts agree and their imaginary parts agree. Real parts: cos(θ + φ) = cos θ cos φ − sin θ sin φ. A1. Imaginary parts: sin(θ + φ) = sin θ cos φ + cos θ sin φ. A1.

Question 11

m is a positive integer and θ is real. Starting from De Moivre's theorem for the positive power m, show that (cos θ + i sin θ)^−m = cos mθ − i sin mθ, and state why this is De Moivre's theorem for the exponent −m.

Accepted Answer

By De Moivre's theorem for the positive power, (cos θ + i sin θ)^m = cos mθ + i sin mθ, so (cos θ + i sin θ)^−m = 1/(cos mθ + i sin mθ). M1 for the reciprocal of the mth power. Multiply top and bottom by the conjugate cos mθ − i sin mθ. M1. The denominator becomes cos²mθ + sin²mθ = 1. A1 for the denominator being 1. So the value is cos mθ − i sin mθ, and since cosine is even and sine is odd that is cos(−mθ) + i sin(−mθ), which is the theorem's own statement with the exponent −m. A1.

Question 12

Let z = cos θ + i sin θ. Show that 16 cos⁵θ = cos 5θ + 5 cos 3θ + 10 cos θ, and hence evaluate ∫ cos⁵θ dθ from 0 to π/2 exactly.

Accepted Answer

By De Moivre, zⁿ + z⁻ⁿ = 2 cos nθ, so in particular z + 1/z = 2 cos θ. Raise that to the fifth power: (z + 1/z)⁵ = z⁵ + 5z³ + 10z + 10/z + 5/z³ + 1/z⁵.
Pair the terms from the ends inwards. That gives 2 cos 5θ + 10 cos 3θ + 20 cos θ, while the left side is 32 cos⁵θ. Halving gives 16 cos⁵θ = cos 5θ + 5 cos 3θ + 10 cos θ.
Now integrate the right side, which has no powers in it: (1/16)[sin 5θ/5 + 5 sin 3θ/3 + 10 sin θ] from 0 to π/2. At π/2 the three sines are 1, −1 and 1, giving (1/16)(1/5 − 5/3 + 10) = (1/16)(128/15) = 8/15. At 0 everything vanishes. B1 for z + 1/z = 2 cos θ, M1 for raising it to the fifth power, M1 for pairing the terms from the ends inwards, A1 for the printed identity, M1 for integrating the multiple angles, A1 for 8/15.

Question 13

The series C = 1 + (1/3)cos θ + (1/9)cos 2θ + … is summed by writing it as the real part of a geometric series. Explain why this series has a sum to infinity, why 1 + cos θ + cos 2θ + … does not, and what weaker condition a sum of finitely many terms would need instead.

Accepted Answer

Adding the matching sine series turns C + iS into a geometric series with complex common ratio (1/3)e^iθ, whose modulus is 1/3 whatever θ is. A modulus below 1 is exactly the condition a geometric series needs for a sum to infinity, so the sum exists for every θ. B1 for the ratio's modulus and the condition. With coefficients of 1 the ratio is e^iθ, of modulus exactly 1: the terms never shrink, so no sum to infinity exists to quote. B1 for the modulus-1 ratio failing. A finite sum uses the n-term formula, which asks only that the ratio is not equal to 1, here meaning that θ is not a multiple of 2π, so the coefficient-1 series can still be summed to n terms. B1 for the ratio-not-1 condition.

Question 14

n is a positive integer and z = cos(π/n) + i sin(π/n). Show that 1 + z + z² + … + zⁿ⁻¹ = 1 + i cot(π/2n).

Accepted Answer

The left side is a geometric series of n terms with first term 1 and common ratio z, and z ≠ 1 because 0 < π/n ≤ π, so its sum is (zⁿ − 1)/(z − 1). M1 for the n-term sum with the ratio z. By De Moivre, zⁿ = cos π + i sin π = −1, so the sum is −2/(z − 1) = −2/[(cos(π/n) − 1) + i sin(π/n)]. A1. Multiply top and bottom by the conjugate of that denominator. M1. The denominator becomes (cos(π/n) − 1)² + sin²(π/n) = 2 − 2cos(π/n), and the numerator becomes −2[(cos(π/n) − 1) − i sin(π/n)] = 2[1 − cos(π/n)] + 2i sin(π/n). A1 for the real denominator. Dividing, the sum is 1 + i sin(π/n)/[1 − cos(π/n)]. Now the double angle formulae with half-angle π/2n give sin(π/n) = 2 sin(π/2n) cos(π/2n) and 1 − cos(π/n) = 2 sin²(π/2n). M1 for both substitutions. The quotient is therefore cos(π/2n)/sin(π/2n) = cot(π/2n), and the sum is 1 + i cot(π/2n). A1. Checking at n = 2: 1 + i and 1 + i cot(π/4) = 1 + i agree.

Question 15

By expanding (cos θ + i sin θ)⁴ and equating real parts, show that cos 4θ = 8cos⁴θ − 8cos²θ + 1, and hence find the exact value of cos(π/8).

Accepted Answer

Writing c for cos θ and s for sin θ, the binomial expansion gives c⁴ + 4ic³s − 6c²s² − 4ics³ + s⁴, using i² = −1 and i⁴ = 1. M1 for the expansion. By De Moivre the same number is cos 4θ + i sin 4θ, so equating real parts gives cos 4θ = c⁴ − 6c²s² + s⁴. A1. Replace s² by 1 − c²: c⁴ − 6c²(1 − c²) + (1 − c²)². M1 for the substitution. That is c⁴ − 6c² + 6c⁴ + 1 − 2c² + c⁴ = 8cos⁴θ − 8cos²θ + 1. A1. Now put θ = π/8, so 4θ = π/2 and cos 4θ = 0: with u = cos²(π/8), 8u² − 8u + 1 = 0, so u = (8 ± √32)/16 = (2 ± √2)/4. M1 for the quadratic in cos²θ and its roots. Since π/8 < π/4, cos(π/8) > cos(π/4) = √2/2, so u must exceed 1/2 and the root (2 − √2)/4 is rejected. Taking the positive square root, because π/8 is acute, cos(π/8) = √(2 + √2)/2, about 0.9239. A1.