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De Moivre's theorem and trigonometric identities questions
Raise a complex number to a power by raising its length and multiplying its angle. One theorem computes high powers in a line and manufactures trig identities to order.
6 original questions · 22 marks · the de moivre's theorem and trigonometric identities notes · Complex numbers
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State De Moivre's theorem, and explain why a number must be in modulus-argument form before the theorem can be applied.
Worked 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.Use De Moivre's theorem to evaluate (1 − i)6.
Worked 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.Evaluate (√3 + i)5, giving the answer in the form a + bi with exact values.
Worked 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. The accuracy mark goes for exact surds; a decimal such as −27.7 + 16i is not the form asked for.Write down the values of (cos π/12 + i sin π/12)6 and (cos π/12 + i sin π/12)−6.
Worked 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, one mark each, and no working is needed for either.By expanding (cos θ + i sin θ)³ and equating imaginary parts, show that sin 3θ = 3 sin θ − 4 sin³θ.
Worked 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. Two of the marks are for the expansion and one for saying which parts are being equated. Dropping the i from the third term is the usual slip, and it puts 3 cos θ sin²θ into the imaginary part by mistake.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.
Worked 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. Pairing the terms is the step the method mark is for; expanding and stopping loses it.
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.
Turning a power into multiple angles is what makes the integral elementary; integrating cos⁵θ directly needs a substitution instead.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise de moivre's theorem and trigonometric identities one question at a time
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