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AS & A Level Further Mathematics (9231)

Complex numbers Banker

11 questions · about 4% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 22 · Q7 (a, b, c)10 marksSumming series with complex numbers · nth roots & roots of unity

2025

  • 2025 Oct/Nov · Paper 22 · Q25 marksnth roots & roots of unity
  • 2025 Oct/Nov · Paper 22 · Q4 (a)5 marksDe Moivre: multiple angles & powers
  • 2025 May/June · Paper 22 · Q15 marksnth roots & roots of unity
  • 2025 May/June · Paper 22 · Q36 marksDe Moivre: multiple angles & powers

2024

  • 2024 Oct/Nov · Paper 22 · Q4 (a, b)10 marksDe Moivre: multiple angles & powers · Solving polynomials via de Moivre
  • 2024 May/June · Paper 22 · Q15 marksnth roots & roots of unity

2023

  • 2023 Oct/Nov · Paper 22 · Q3 (a, b)8 marksDe Moivre: multiple angles & powers · Solving polynomials via de Moivre
  • 2023 May/June · Paper 22 · Q3 (a)5 marksDe Moivre: multiple angles & powers

2022

  • 2022 Oct/Nov · Paper 22 · Q5 (a, b, c)10 marksDe Moivre: multiple angles & powers · Solving polynomials via de Moivre · nth roots & roots of unity
  • 2022 May/June · Paper 22 · Q7 (a, b)11 marksDe Moivre: multiple angles & powers · Solving polynomials via de Moivre

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1De Moivre: multiple angles & powers · Core · Paper 2 style · 5 marks
  1. (a) Express 1 + i√3 in the form r e^(iθ), where r > 0 and −π < θ ≤ π. [2]
  2. (b) Hence use de Moivre's theorem to find (1 + i√3)⁸ in the form a + ib, where a and b are exact. [3]
Show answer
  1. (a) 2e^(iπ/3)
    B1 r = 2; B1 θ = π/3.
  2. (b) −128 + 128√3 i
    M1 2⁸ e^(8iπ/3); M1 8π/3 = 2π + 2π/3; A1 256(−½ + (√3/2)i) = −128 + 128√3 i.
Q2De Moivre: multiple angles & powers · Core · Paper 2 style · 4 marks
  1. (a) Use de Moivre's theorem to find (√3 − i)⁻⁴ in the form a + ib, giving a and b exactly. [4]
Show answer
  1. (a) −1/32 + (√3/32)i
    B1 √3 − i = 2e^(−iπ/6); M1 power −4: 2⁻⁴ e^(2iπ/3); M1 cos(2π/3) + i sin(2π/3) = −½ + (√3/2)i; A1 −1/32 + (√3/32)i.
Q3De Moivre: multiple angles & powers · Core · Paper 2 style · 7 marks
  1. (a) Use de Moivre's theorem to show that sin 3θ = 3sin θ − 4sin³θ. [4]
  2. (b) Hence find the exact value of ∫ sin³θ dθ with limits 0 and ½π. [3]
Show answer
  1. (a) Shown
    M1 (cos θ + i sin θ)³ = cos 3θ + i sin 3θ; M1 expand, imaginary part 3cos²θ sin θ − sin³θ; M1 cos²θ = 1 − sin²θ; A1 3sin θ − 4sin³θ (AG).
  2. (b) 2/3
    M1 sin³θ = ¼(3sin θ − sin 3θ); M1 [¼(−3cos θ + ⅓cos 3θ)] from 0 to ½π; A1 ¼(3 − ⅓) = 2/3.
Q4De Moivre: multiple angles & powers · Core · Paper 2 style · 7 marks
  1. (a) By writing z = cos θ + i sin θ and using z + 1/z = 2cos θ, show that cos⁴θ = ⅛(cos 4θ + 4cos 2θ + 3). [4]
  2. (b) Hence find the exact value of ∫ cos⁴θ dθ with limits 0 and ¼π. [3]
Show answer
  1. (a) Shown
    M1 (z + 1/z)⁴ = 16cos⁴θ; M1 expand: z⁴ + z⁻⁴ + 4(z² + z⁻²) + 6; M1 zⁿ + z⁻ⁿ = 2cos nθ; A1 result (AG).
  2. (b) ¼ + 3π/32
    M1 [⅛(¼sin 4θ + 2sin 2θ + 3θ)]; M1 limits; A1 ⅛(2 + 3π/4) = ¼ + 3π/32.
Q5De Moivre: multiple angles & powers · Standard · Paper 2 style · 6 marks
  1. (a) Use de Moivre's theorem to show that cos 5θ = 16cos⁵θ − 20cos³θ + 5cos θ. [5]
  2. (b) Hence find the exact value of 16cos⁵(π/10) − 20cos³(π/10) + 5cos(π/10). [1]
Show answer
  1. (a) Shown
    M1 real part of (c + is)⁵; A1 c⁵ − 10c³s² + 5cs⁴; M1 s² = 1 − c²; M1 expand (1 − c²)²; A1 16c⁵ − 20c³ + 5c (AG).
  2. (b) 0
    B1 cos(5π/10) = cos(π/2) = 0.
Q6De Moivre: multiple angles & powers · Standard · Paper 2 style · 9 marks
  1. (a) Use de Moivre's theorem to show that tan 3θ = (3t − t³)/(1 − 3t²), where t = tan θ. [4]
  2. (b) By considering tan 3θ = 1, show that t = tan(π/12) is a root of t³ − 3t² − 3t + 1 = 0. [2]
  3. (c) Hence find the exact value of tan(π/12). [3]
Show answer
  1. (a) Shown
    M1 cos 3θ = c³ − 3cs², sin 3θ = 3c²s − s³; M1 divide; M1 by c³; A1 (AG).
  2. (b) Shown
    M1 3t − t³ = 1 − 3t²; A1 rearranged, and tan(3 × π/12) = 1 (AG).
  3. (c) 2 − √3
    M1 t = −1 is a root (tan(3π/4) = −1), factor (t + 1)(t² − 4t + 1); M1 t = 2 ± √3; A1 tan(π/12) is the smallest positive root, 2 − √3.
Q7De Moivre: multiple angles & powers · Standard · Paper 2 style · 8 marks
  1. (a) Using z = cos θ + i sin θ and z − 1/z = 2i sin θ, show that sin⁵θ = (1/16)(sin 5θ − 5sin 3θ + 10sin θ). [5]
  2. (b) Hence find the exact value of ∫ sin⁵θ dθ with limits 0 and ½π. [3]
Show answer
  1. (a) Shown
    M1 (z − 1/z)⁵ = 32i sin⁵θ; M1 expand: (z⁵ − z⁻⁵) − 5(z³ − z⁻³) + 10(z − z⁻¹); M1 zⁿ − z⁻ⁿ = 2i sin nθ; A1 32i sin⁵θ = 2i(sin 5θ − 5sin 3θ + 10sin θ); A1 (AG).
  2. (b) 8/15
    M1 integrate each term; M1 values 1/5, 1/3, 1; A1 (1/16)(1/5 − 5/3 + 10) = 8/15.
Q8De Moivre: multiple angles & powers · Standard · Paper 2 style · 4 marks
  1. (a) Find the smallest positive integer n for which (√3 + i)ⁿ is a negative real number, and state the value of (√3 + i)ⁿ for this n. [4]
Show answer
  1. (a) n = 6; (√3 + i)⁶ = −64
    B1 √3 + i = 2e^(iπ/6); M1 nπ/6 = π + 2kπ; A1 n = 6; A1 2⁶e^(iπ) = −64.
Q9De Moivre: multiple angles & powers · Challenge · Paper 2 style · 9 marks
  1. (a) Use de Moivre's theorem to show that sin 6θ / sin θ = 32cos⁵θ − 32cos³θ + 6cos θ, for sin θ ≠ 0. [6]
  2. (b) Hence find the values of cos θ, for 0 < θ < π, which satisfy sin 6θ = 6cos θ sin θ, giving exact answers. [3]
Show answer
  1. (a) Shown
    M1 Im(c + is)⁶ = 6c⁵s − 20c³s³ + 6cs⁵; M1 divide by s; M1 s² = 1 − c², s⁴ = (1 − c²)²; M1 expand; A1 collect: 6c⁵ − 20c³ + 20c⁵ + 6c − 12c³ + 6c⁵; A1 32c⁵ − 32c³ + 6c (AG).
  2. (b) cos θ = 0 only (θ = ½π)
    M1 32c⁵ − 32c³ = 0; M1 c = 0 or c² = 1; A1 c = ±1 gives sin θ = 0, not in range, so cos θ = 0 (θ = ½π).
Q10De Moivre: multiple angles & powers · Challenge · Paper 2 style · 8 marks
  1. (a) Show that cos⁵θ = (1/16)(cos 5θ + 5cos 3θ + 10cos θ). [4]
  2. (b) Hence find the exact value of ∫ cos⁵θ dθ with limits 0 and ⅓π. [4]
Show answer
  1. (a) Shown
    M1 (z + 1/z)⁵ = 32cos⁵θ; M1 expand binomially; M1 pair terms as 2cos nθ; A1 (AG).
  2. (b) 49√3/160
    M1 (1/16)[⅕sin 5θ + (5/3)sin 3θ + 10sin θ]; M1 sin(5π/3) = −√3/2, sin π = 0, sin(π/3) = √3/2; M1 (1/16)(−√3/10 + 5√3); A1 49√3/160.
Q11De Moivre: multiple angles & powers · Challenge · Paper 2 style · 8 marks
  1. (a) Use de Moivre's theorem to show that cos 4θ = 8cos⁴θ − 8cos²θ + 1. [4]
  2. (b) By considering the equation cos 4θ = 0, show that cos(π/8) = ½√(2 + √2). [4]
Show answer
  1. (a) Shown
    M1 Re(c + is)⁴ = c⁴ − 6c²s² + s⁴; M1 s² = 1 − c²; A1 c⁴ − 6c² + 6c⁴ + 1 − 2c² + c⁴; A1 (AG).
  2. (b) Shown
    M1 8c⁴ − 8c² + 1 = 0 with roots cos(π/8), cos(3π/8), …; M1 c² = (2 ± √2)/4; A1 cos(π/8) is the largest positive root so take +; A1 c = ½√(2 + √2) (AG).
Q12De Moivre: multiple angles & powers · Challenge · Paper 2 style · 6 marks
  1. (a) Show that 1 + cos 2θ + i sin 2θ = 2cos θ (cos θ + i sin θ). [2]
  2. (b) Hence show that (1 + cos 2θ + i sin 2θ)ⁿ = 2ⁿcosⁿθ (cos nθ + i sin nθ), and find the exact real part of (1 + cos 2θ + i sin 2θ)⁴ when θ = π/6. [4]
Show answer
  1. (a) Shown
    M1 1 + cos 2θ = 2cos²θ, sin 2θ = 2sin θ cos θ; A1 factorise (AG).
  2. (b) −9/2
    M1 de Moivre applied to (cos θ + i sin θ)ⁿ; A1 shown; M1 16(√3/2)⁴ cos(2π/3); A1 16 × 9/16 × (−½) = −9/2.
Q13De Moivre: multiple angles & powers · Challenge · Paper 2 style · 8 marks
  1. (a) Use de Moivre's theorem to show that sin²θ cos⁴θ = (1/32)(2 + cos 2θ − 2cos 4θ − cos 6θ). [6]
  2. (b) Hence find the exact value of ∫ sin²θ cos⁴θ dθ with limits 0 and ½π. [2]
Show answer
  1. (a) Shown
    M1 (z + 1/z)⁴(z − 1/z)² = (2cos θ)⁴(2i sin θ)² = −64 sin²θ cos⁴θ; M1 (z + 1/z)²(z² − 1/z²)²; M1 expand: z⁶ + z⁻⁶ + 2(z⁴ + z⁻⁴) − (z² + z⁻²) − 4; M1 = 2cos 6θ + 4cos 4θ − 2cos 2θ − 4; A1 divide by −64; A1 (AG).
  2. (b) π/32
    M1 the cosine terms integrate to 0 over [0, ½π]; A1 (1/32)(2 × ½π) = π/32.
Q14Solving polynomials via de Moivre · Core · Paper 2 style · 6 marks
  1. (a) Given that cos 3θ = 4cos³θ − 3cos θ, show that the substitution x = cos θ turns the equation 8x³ − 6x − 1 = 0 into cos 3θ = ½. [2]
  2. (b) Hence find the three roots of 8x³ − 6x − 1 = 0, giving each in the form cos(kπ) and correct to 3 decimal places. [4]
Show answer
  1. (a) Shown
    M1 8x³ − 6x = 2(4cos³θ − 3cos θ) = 2cos 3θ; A1 2cos 3θ = 1 (AG).
  2. (b) cos(π/9) = 0.940, cos(5π/9) = −0.174, cos(7π/9) = −0.766
    M1 3θ = π/3, 5π/3, 7π/3; A1 θ = π/9, 5π/9, 7π/9; M1 x = cos θ; A1 three values correct to 3 d.p.
Q15Solving polynomials via de Moivre · Core · Paper 2 style · 6 marks
  1. (a) Given that sin 3θ = 3sin θ − 4sin³θ, show that the equation 8x³ − 6x + √2 = 0 becomes sin 3θ = √2/2 when x = sin θ. [2]
  2. (b) Hence find the three roots of 8x³ − 6x + √2 = 0, giving exact forms sin(kπ) and values correct to 4 decimal places. [4]
Show answer
  1. (a) Shown
    M1 8x³ − 6x = −2(3sin θ − 4sin³θ) = −2sin 3θ; A1 −2sin 3θ + √2 = 0 (AG).
  2. (b) sin(π/12) = 0.2588, sin(π/4) = 0.7071, sin(17π/12) = −0.9659
    M1 3θ = π/4, 3π/4, 9π/4, …; M1 θ = π/12, π/4, 3π/4, 11π/12, 17π/12, 19π/12 give only three distinct values of sin θ; A1 two roots; A1 third root (sum of roots = 0 as a check).
Q16Solving polynomials via de Moivre · Standard · Paper 2 style · 8 marks
  1. (a) Use de Moivre's theorem to show that tan 4θ = (4t − 4t³)/(1 − 6t² + t⁴), where t = tan θ. [4]
  2. (b) Hence solve t⁴ + 4t³ − 6t² − 4t + 1 = 0, giving the roots in the form tan(kπ) and correct to 3 decimal places. [4]
Show answer
  1. (a) Shown
    M1 cos 4θ = c⁴ − 6c²s² + s⁴; M1 sin 4θ = 4c³s − 4cs³; M1 divide by c⁴; A1 (AG).
  2. (b) tan(π/16) = 0.199, tan(5π/16) = 1.497, tan(9π/16) = −5.027, tan(13π/16) = −0.668
    M1 equation ⇔ 4t − 4t³ = 1 − 6t² + t⁴ ⇔ tan 4θ = 1; M1 4θ = π/4 + kπ; A1 θ = π/16, 5π/16, 9π/16, 13π/16; A1 decimals.
Q17Solving polynomials via de Moivre · Standard · Paper 2 style · 7 marks
  1. (a) Given that cos 5θ = 16cos⁵θ − 20cos³θ + 5cos θ, show that the non-zero roots of 16x⁴ − 20x² + 5 = 0 are cos(π/10), cos(3π/10), cos(7π/10) and cos(9π/10). [4]
  2. (b) Hence show that cos(π/10) cos(3π/10) = √5/4. [3]
Show answer
  1. (a) Shown
    M1 cos 5θ = 0 gives x(16x⁴ − 20x² + 5) = 0 with x = cos θ; M1 5θ = π/2, 3π/2, 7π/2, 9π/2; A1 θ = π/10, 3π/10, 7π/10, 9π/10, distinct non-zero cosines; A1 x = 0 corresponds to θ = π/2 (AG).
  2. (b) Shown
    M1 cos²(π/10) and cos²(3π/10) are the roots of 16y² − 20y + 5 = 0; M1 product of roots 5/16; A1 both cosines positive, product √5/4 (AG).
Q18Solving polynomials via de Moivre · Standard · Paper 2 style · 6 marks
  1. (a) By substituting x = 2cos θ and using cos 3θ = 4cos³θ − 3cos θ, solve the equation x³ − 3x + 1 = 0. Give each root in exact trigonometric form and correct to 3 significant figures. [6]
Show answer
  1. (a) 2cos(2π/9) = 1.53, 2cos(4π/9) = 0.347, 2cos(8π/9) = −1.88
    M1 8cos³θ − 6cos θ + 1 = 0; M1 2cos 3θ = −1; A1 cos 3θ = −½; M1 3θ = 2π/3, 4π/3, 8π/3; A1 θ = 2π/9, 4π/9, 8π/9; A1 values.
Q19Solving polynomials via de Moivre · Challenge · Paper 2 style · 7 marks
  1. (a) Using cos 5θ = 16cos⁵θ − 20cos³θ + 5cos θ, show that x = ½ is a root of 32x⁵ − 40x³ + 10x − 1 = 0 and find, in the form cos(kπ), the other four roots. [5]
  2. (b) Hence find the exact value of cos(π/15) + cos(7π/15) + cos(11π/15) + cos(13π/15). [2]
Show answer
  1. (a) cos(π/15), cos(7π/15), cos(11π/15), cos(13π/15)
    M1 equation ⇔ 2cos 5θ = 1; M1 5θ = ±π/3 + 2nπ; A1 x = cos(π/3) = ½ (from 5θ = 5π/3); M1 θ = π/15, 7π/15, 11π/15, 13π/15; A1 all four.
  2. (b) −½
    M1 sum of the five roots = 0 (no x⁴ term); A1 0 − ½ = −½.
Q20Solving polynomials via de Moivre · Challenge · Paper 2 style · 8 marks
  1. (a) Use de Moivre's theorem to show that sin 5θ = 16sin⁵θ − 20sin³θ + 5sin θ. [4]
  2. (b) Hence show that sin²(π/5) and sin²(2π/5) are the roots of 16y² − 20y + 5 = 0, and find the exact value of sin²(π/5). [4]
Show answer
  1. (a) Shown
    M1 Im(c + is)⁵ = 5c⁴s − 10c²s³ + s⁵; M1 c² = 1 − s²; A1 expand; A1 (AG).
  2. (b) sin²(π/5) = (5 − √5)/8
    M1 sin 5θ = 0, θ = π/5, 2π/5 give non-zero s with 16s⁴ − 20s² + 5 = 0; A1 y = s²; M1 y = (5 ± √5)/8; A1 sin²(π/5) < sin²(2π/5), so (5 − √5)/8.
Q21nth roots & roots of unity · Core · Paper 2 style · 5 marks
  1. (a) Solve the equation z⁵ = 1, giving the roots in the form e^(iθ), where −π < θ ≤ π. [3]
  2. (b) Show that the sum of the five roots is zero. [2]
Show answer
  1. (a) 1, e^(±2πi/5), e^(±4πi/5)
    M1 z = e^(2kπi/5); A1 k = 0, ±1; A1 k = ±2.
  2. (b) Shown
    M1 roots form a geometric series 1 + ω + ω² + ω³ + ω⁴ = (1 − ω⁵)/(1 − ω), ω = e^(2πi/5); A1 ω⁵ = 1, ω ≠ 1, so sum 0 (AG).
Q22nth roots & roots of unity · Core · Paper 2 style · 5 marks
  1. (a) Find the three roots of z³ = 8i, giving them in the form r e^(iθ) with −π < θ ≤ π, and in the form x + iy. [5]
Show answer
  1. (a) 2e^(iπ/6) = √3 + i; 2e^(5iπ/6) = −√3 + i; 2e^(−iπ/2) = −2i
    B1 8i = 8e^(iπ/2); M1 r = 2, θ = π/6 + 2kπ/3; A1 three arguments; A1 two in x + iy form; A1 third.
Q23nth roots & roots of unity · Core · Paper 2 style · 5 marks
  1. (a) The complex number ω is a root of z³ = 1 with ω ≠ 1. Show that 1 + ω + ω² = 0. [2]
  2. (b) Find the values of (1 + ω)(1 + ω²) and (2 + ω)(2 + ω²). [3]
Show answer
  1. (a) Shown
    M1 z³ − 1 = (z − 1)(z² + z + 1); A1 ω ≠ 1 so ω² + ω + 1 = 0 (AG).
  2. (b) 1 and 3
    M1 expand (1 + ω)(1 + ω²) = 1 + ω + ω² + ω³; A1 = 0 + 1 = 1; A1 (2 + ω)(2 + ω²) = 4 + 2(ω + ω²) + ω³ = 4 − 2 + 1 = 3.
Q24nth roots & roots of unity · Standard · Paper 2 style · 6 marks
  1. (a) Express −8 + 8√3 i in the form r e^(iθ), where r > 0 and −π < θ ≤ π. [2]
  2. (b) Hence find the four roots of z⁴ = −8 + 8√3 i, giving your answers exactly in the form x + iy. [4]
Show answer
  1. (a) 16e^(2πi/3)
    B1 r = 16; B1 θ = 2π/3.
  2. (b) √3 + i, −1 + √3 i, −√3 − i, 1 − √3 i
    M1 r = 2, θ = π/6 + kπ/2; A1 arguments π/6, 2π/3, −5π/6, −π/3; A1 two roots in x + iy form; A1 all four.
Q25nth roots & roots of unity · Standard · Paper 2 style · 7 marks
  1. (a) Find the cube roots of 4√2(1 + i), giving your answers in the form r e^(iθ) with −π < θ ≤ π. [4]
  2. (b) The roots are represented by points A, B and C in an Argand diagram. Explain why triangle ABC is equilateral and find its exact area. [3]
Show answer
  1. (a) 2e^(iπ/12), 2e^(3πi/4), 2e^(−7πi/12)
    B1 4√2(1 + i) = 8e^(iπ/4); M1 r = 2; M1 θ = π/12 + 2kπ/3; A1 all three.
  2. (b) 3√3
    B1 points on a circle of radius 2 with arguments differing by 2π/3; M1 side = 2 × 2 sin(π/3) = 2√3, area = (√3/4)(2√3)²; A1 3√3.
Q26nth roots & roots of unity · Challenge · Paper 2 style · 8 marks
  1. (a) Show that the equation (z + 1)⁵ = z⁵ can be written as ((z + 1)/z)⁵ = 1 and explain why it has exactly four roots. [2]
  2. (b) Show that the roots are z = 1/(ω − 1), where ω = e^(2kπi/5), k = 1, 2, 3, 4, and deduce that each root has real part −½. [4]
  3. (c) Hence write down the roots in the form −½ + iy, giving y correct to 3 significant figures. [2]
Show answer
  1. (a) Shown
    B1 z = 0 is not a root, so divide by z⁵; B1 (z + 1)/z = ω, a fifth root of unity with ω ≠ 1 (ω = 1 gives 1 = 0); the equation is quartic, four roots.
  2. (b) Re z = −½
    M1 z + 1 = ωz so z = 1/(ω − 1); M1 ω − 1 = e^(iφ) − 1 = 2i sin(φ/2) e^(iφ/2) with φ = 2kπ/5; M1 1/(ω − 1) = −(i/2)e^(−iφ/2)/sin(φ/2) = −½ − (i/2)cot(φ/2); A1 real part −½ (AG).
  3. (c) −½ ± 0.688i, −½ ± 0.162i
    M1 y = −½cot(kπ/5); A1 ±0.688 and ±0.162.
Q27nth roots & roots of unity · Challenge · Paper 2 style · 7 marks
  1. (a) A regular hexagon in an Argand diagram has its centre at the origin and one vertex at the point representing 3 + i. Find the complex numbers represented by the other five vertices, in the form x + iy with exact surds. [4]
  2. (b) Show that the vertices are the roots of z⁶ = −352 + 936i. [3]
Show answer
  1. (a) (3 − √3)/2 + ((1 + 3√3)/2)i, (−3 − √3)/2 + ((3√3 − 1)/2)i, −3 − i, (−3 + √3)/2 − ((1 + 3√3)/2)i, (3 + √3)/2 + ((1 − 3√3)/2)i
    M1 multiply by e^(iπ/3) = ½ + (√3/2)i repeatedly; A1 first new vertex; A1 −3 − i (rotation by π); A1 all.
  2. (b) Shown
    M1 each vertex is (3 + i)e^(kπi/3), so z⁶ = (3 + i)⁶; M1 (3 + i)² = 8 + 6i, (3 + i)³ = 18 + 26i; A1 (18 + 26i)² = −352 + 936i (AG).
Q28Summing series with complex numbers · Core · Paper 2 style · 7 marks
  1. (a) Let z = cos θ + i sin θ. By considering the infinite geometric series 1 + ½z + ¼z² + ⅛z³ + …, show that 1 + ½cos θ + ¼cos 2θ + ⅛cos 3θ + … = (4 − 2cos θ)/(5 − 4cos θ). [6]
  2. (b) Hence find the exact value of the sum when θ = ½π. [1]
Show answer
  1. (a) Shown
    M1 |z/2| < 1, sum = 1/(1 − ½z) = 2/(2 − z); M1 de Moivre: real part of the series is the required sum; M1 multiply by conjugate 2 − cos θ + i sin θ; M1 denominator (2 − cos θ)² + sin²θ = 5 − 4cos θ; A1 real part 2(2 − cos θ)/(5 − 4cos θ); A1 (AG).
  2. (b) 4/5
    B1 (4 − 0)/(5 − 0) = 4/5.
Q29Summing series with complex numbers · Standard · Paper 2 style · 8 marks
  1. (a) By considering the sum of a geometric series in z = e^(iθ), show that, for sin θ ≠ 0, Σ (r = 1 to n) cos(2r − 1)θ = sin 2nθ / (2sin θ). [6]
  2. (b) Hence find the exact value of cos(π/7) + cos(3π/7) + cos(5π/7). [2]
Show answer
  1. (a) Shown
    M1 Σ e^((2r − 1)iθ) = e^(iθ)(1 − e^(2niθ))/(1 − e^(2iθ)); M1 = (e^(2niθ) − 1)/(e^(iθ) − e^(−iθ)); M1 denominator 2i sin θ; M1 real part of (cos 2nθ − 1 + i sin 2nθ)/(2i sin θ); A1 sin 2nθ/(2sin θ); A1 (AG) with sum of real parts = sum of cosines.
  2. (b) ½
    M1 n = 3, θ = π/7: sin(6π/7)/(2sin(π/7)); A1 sin(6π/7) = sin(π/7), so ½.
Q30Summing series with complex numbers · Challenge · Paper 2 style · 8 marks
  1. (a) Show that 1 + e^(iθ) = 2cos(½θ) e^(iθ/2). [2]
  2. (b) By considering (1 + e^(iθ))ⁿ and the binomial theorem, show that Σ (r = 0 to n) C(n, r) sin rθ = 2ⁿ cosⁿ(½θ) sin(½nθ), where C(n, r) is the binomial coefficient. [4]
  3. (c) Hence find the exact value of Σ (r = 0 to 6) C(6, r) sin(rπ/3). [2]
Show answer
  1. (a) Shown
    M1 e^(iθ/2)(e^(−iθ/2) + e^(iθ/2)); A1 = 2cos(½θ)e^(iθ/2) (AG).
  2. (b) Shown
    M1 (1 + e^(iθ))ⁿ = Σ C(n, r)e^(irθ); M1 = 2ⁿcosⁿ(½θ)e^(inθ/2) by de Moivre; M1 take imaginary parts; A1 (AG).
  3. (c) 0
    M1 2⁶cos⁶(π/6) sin(π); A1 0.

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

De Moivre: multiple angles & powers 7 questions · 7/9 sessions
Solving polynomials via de Moivre 4 questions · 4/9 sessions
nth roots & roots of unity 5 questions · 5/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
De Moivre: multiple angles & powers 367/9
Solving polynomials via de Moivre 184/9
nth roots & roots of unity 185/9
Summing series with complex numbers Rare81/9

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