Complex numbers — AS & A Level Further Mathematics (9231) questions by topic

De Moivre's theorem and root-finding define this topic: finding the roots of z^3 = 27 - 27i in modulus-argument form, using a binomial expansion of (z + 1/z)^7 to express cos^7 theta as a sum of multiple-angle cosines, and showing an expansion for sin 7theta.

Find and show are the two top command words across 17 questions from 2023 to 2025 papers, worth from 5 to 14 marks, where converting cleanly between exponential and multiple-angle cosine or sine forms is the main technical hurdle. With no examiner-report extracts logged for this topic, submitting each root-finding or expansion derivation for instant marking is the way to confirm the argument range and coefficients are exactly right before the final expression is trusted.

May–June 2025, Paper 21

Question 1 · 5 marks

Find the roots of the equation z^3 = 27 - 27i, giving your answers in the form r e^(itheta), where r 0 and -pi < theta <= pi.

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Question 3 · 6 marks

By considering the binomial expansion of (z - 1/z)^5, where z = cos(theta) + i sin(theta), use de Moivre's theorem to show that cosec^5(theta) = a / (sin(5theta) + b sin(3theta) +…

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May–June 2025, Paper 22

Question 1 · 5 marks

Find the roots of the equation z^3 = 27 - 27i, giving your answers in the form re^(itheta), where r 0 and -pi < theta <= pi.

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Question 3 · 6 marks

By considering the binomial expansion of (z - 1/z)^5, where z = cos(theta) + isin(theta), use de Moivre's theorem to show that cosec^5(theta) = a / (sin(5theta) + bsin(3theta) +…

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May–June 2025, Paper 23

Question 5 · 10 marks

Use de Moivre's theorem to show that sec(5theta) = sec^5(theta) / (5sec^4(theta) - 20sec^2(theta) + 16).

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May–June 2025, Paper 24

Question 5 · 8 marks

Use de Moivre's theorem to show that sin 7θ = -64 sin^7 θ + 112 sin^5 θ - 56 sin^3 θ + 7 sin θ.

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May–June 2024, Paper 21

Question 1 · 5 marks

Find the roots of the equation z^3 = -108sqrt(3) + 108i, giving your answers in the form r(cos(theta) + i sin(theta)), where r 0 and 0 < theta < 2pi.

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May–June 2024, Paper 22

Question 1 · 5 marks

Find the roots of the equation z^3 = -108sqrt(3) + 108i, giving your answers in the form r(cos(theta) + isin(theta)), where r 0 and 0 < theta < 2pi.

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October–November 2024, Paper 21

Question 8 · 14 marks

By considering the binomial expansion of (z + 1/z)^7, where z = cos theta + i sin theta, use de Moivre's theorem to show that cos^7 theta = a cos 7 theta + b cos 5 theta + c cos 3…

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October–November 2024, Paper 22

Question 4 · 10 marks

Use de Moivre's theorem to show that cot(6 theta) = (cot^6(theta) - 15 cot^4(theta) + 15 cot^2(theta) - 1) / (6 cot^5(theta) - 20 cot^3(theta) + 6 cot(theta)).

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October–November 2024, Paper 23

Question 8 · 14 marks

By considering the binomial expansion of (z + 1/z)^7, where z = cos(theta) + i sin(theta), use de Moivre's theorem to show that cos^7(theta) = a cos(7 theta) + b cos(5 theta) + c…

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May–June 2023, Paper 21

Question 3 · 8 marks

By considering the binomial expansion of (z + z^(-1))^4, where z = cos(theta) + i sin(theta), use de Moivre's theorem to show that cos^4(theta) = (1/8)(cos(4theta) + 4 cos(2theta)…

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May–June 2023, Paper 22

Question 3 · 8 marks

By considering the binomial expansion of (z + z^(-1))^4, where z = cos(theta) + isin(theta), use de Moivre's theorem to show that cos^4(theta) = (1/8)(cos(4theta) + 4cos(2theta) +…

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May–June 2023, Paper 23

Question 3 · 7 marks

By considering the binomial expansions of (z + 1/z)^4 and (z - 1/z)^4, where z = cos(theta) + i sin(theta), use de Moivre's theorem to show that cot^4(theta) = (cos 4theta + a cos…

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October–November 2023, Paper 21

Question 2 · 5 marks

Find the roots of the equation (z + 5i)^3 = 4 + 4 sqrt(3) i, giving your answers in the form r cos(theta) + i(r sin(theta) - 5), where r 0 and 0 < theta < 2pi.

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October–November 2023, Paper 22

Question 3 · 8 marks

Use de Moivre's theorem to show that cos 5theta = 16 cos^5 theta - 20 cos^3 theta + 5 cos theta.

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October–November 2023, Paper 23

Question 2 · 5 marks

Find the roots of the equation (z + 5i)^3 = 4 + 4 sqrt(3) i, giving your answers in the form r cos(theta) + i(r sin(theta) - 5), where r 0 and 0 < theta < 2pi.

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