All subjects · Further Mathematics · WFM02 Further Pure Mathematics 2

De Moivre's theorem

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

De Moivre's theorem says that raising cos θ + i sin θ to a power n multiplies the angle by n. It turns long binomial expansions into one line of working.

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The theorem

(cos θ + i sin θ)n = cos nθ + i sin nθ, for any integer n.

In exponential form this is just (eiθ)n = einθ. For z = r(cos θ + i sin θ): zn = rn(cos nθ + i sin nθ).

Proof by induction (positive n)

Basis: n = 1 is true.

Assume (cos θ + i sin θ)k = cos kθ + i sin kθ. Then

(cos θ + i sin θ)k + 1 = (cos kθ + i sin kθ)(cos θ + i sin θ) = (cos kθ cos θ - sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ) = cos(k + 1)θ + i sin(k + 1)θ, using the addition formulae.

So true for n = k implies true for n = k + 1; true for n = 1, so true for all positive integers n. You must be able to write this proof.

Negative powers

For n = -m with m positive: (cos θ + i sin θ)-m = 1/(cos mθ + i sin mθ) = cos mθ - i sin mθ = cos(-mθ) + i sin(-mθ). So the theorem holds for negative integers too.

Working with powers

Convert to modulus-argument form, apply the theorem, then evaluate fully: the scheme of work warns not to leave answers like 64(cos 2π + i sin 2π). If cos and sin appear with a minus sign, such as cos θ - i sin θ, rewrite it as cos(-θ) + i sin(-θ) first.

Worked example

Find (1 + i√3)6.

  1. |1 + i√3| = 2 and arg = π/3.
  2. (1 + i√3)6 = 26(cos 2π + i sin 2π) by de Moivre's theorem.
  3. cos 2π = 1 and sin 2π = 0.

Answer: 64

Key idea: (r eiθ)n = rn einθ: raise the modulus to the power n and multiply the argument by n. Always evaluate the final answer.

Check yourself

  1. Find the imaginary part of (√3 + i)5.
    Show answer

    16. √3 + i = 2eiπ/6. So (√3 + i)5 = 32(cos(5π/6) + i sin(5π/6)) and the imaginary part is 32 × 1/2 = 16.

  2. Find (1 - i)8.
    Show answer

    16. 1 - i = √2 e-iπ/4. (1 - i)8 = (√2)8 e-2πi = 16.

  3. Simplify (cos 3θ + i sin 3θ)/(cos θ - i sin θ)2.
    • cos θ + i sin θ
    • cos 5θ + i sin 5θ
    • cos 5θ - i sin 5θ
    • cos 6θ + i sin 6θ
    Show answer

    cos 5θ + i sin 5θ. cos θ - i sin θ = e-iθ, so the denominator is e-2iθ. e3iθ/e-2iθ = e5iθ.

  4. In the inductive step of the proof of de Moivre's theorem, which identity is used to simplify cos kθ cos θ - sin kθ sin θ?
    • cos 2A = 2cos2 A - 1
    • sin(A + B) = sin A cos B + cos A sin B
    • cos(A + B) = cos A cos B - sin A sin B
    • cos2 A + sin2 A = 1
    Show answer

    cos(A + B) = cos A cos B - sin A sin B. cos kθ cos θ - sin kθ sin θ = cos(kθ + θ) = cos(k + 1)θ.

  5. Find the real part of (cos(π/9) + i sin(π/9))6.
    Show answer

    -0.5. By de Moivre: cos(6π/9) + i sin(6π/9) = cos(2π/3) + i sin(2π/3). The real part is -1/2.

  6. Which expression equals (cos θ + i sin θ)-n?
    • -cos nθ - i sin nθ
    • 1/cos nθ + i/sin nθ
    • cos nθ - i sin nθ
    • cos(-nθ) - i sin(-nθ)
    Show answer

    cos nθ - i sin nθ. (eiθ)-n = e-inθ = cos nθ - i sin nθ, because cos is even and sin is odd.

  7. Find the smallest positive integer n for which (1 + i)n is real and positive.
    Show answer

    8. arg(1 + i)n = nπ/4. This must be a multiple of 2π, so n = 8 (and (1 + i)8 = 16).

  8. Find the modulus of (2 - 2i)4.
    Show answer

    64. |2 - 2i| = 2√2, and (2√2)4 = 64.

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Last updated 8 November 2026. Spotted a mistake? Tell us.