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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9231Paper 2 · Further Pure Mathematics 2Complex numbersDe Moivre's theorem and roots of unity
Complex numbers · Subtopic 1 of 1

De Moivre's Theorem and Roots of Unity

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta: to raise a complex number to a power, raise the modulus and multiply the argument. Run it backwards to find all nn nnth roots, which sit evenly round a circle, and expand it to get identities for cos⁡nθ\cos n\theta and sin⁡nθ\sin n\theta.

Cambridge 9231 Paper 2≈ 50 minRevision 1 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • State and use de Moivre's theorem for integer powers.
  • Express cos⁡nθ\cos n\theta and sin⁡nθ\sin n\theta in powers of cos⁡θ\cos\theta, sin⁡θ\sin\theta, and the reverse.
  • Find the nnth roots of a complex number and the roots of unity.
Pause and think

Multiplying by ii rotates a point 90∘90^\circ. What does multiplying by ii four times do?

Section 2 of 14

Core idea

Start hereSpinning a point round a circle

Multiplying by ii turns a point a quarter-turn about the origin. Multiplying by cos⁡θ+isin⁡θ\cos\theta+i\sin\theta turns it by θ\theta. So multiplying by it nn times turns it by nθn\theta. That simple picture is de Moivre's theorem, and it unlocks powers, roots and trig identities in one go.

The big idea

(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta, i.e. (reiθ)n=rneinθ(re^{i\theta})^n=r^ne^{in\theta}. The equation zn=wz^n=w has nn roots, equally spaced on a circle.

1. The theorem

Proof for positive integers (induction)
  1. n=1: cos⁡θ+isin⁡θn=1:\ \cos\theta+i\sin\theta
    Trivially true.
  2. (cos⁡θ+isin⁡θ)k+1=(cos⁡kθ+isin⁡kθ)(cos⁡θ+isin⁡θ)(\cos\theta+i\sin\theta)^{k+1}=(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)
    Assume true for n=kn=k.
  3. =(cos⁡kθcos⁡θ−sin⁡kθsin⁡θ)+i(sin⁡kθcos⁡θ+cos⁡kθsin⁡θ)=(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)
    Expand; i2=−1i^2=-1.
  4. =cos⁡(k+1)θ+isin⁡(k+1)θ=\cos(k+1)\theta+i\sin(k+1)\theta
    Compound-angle formulae.
True for all positive integers; negative nn follows from 1cos⁡θ+isin⁡θ=cos⁡θ−isin⁡θ\frac1{\cos\theta+i\sin\theta}=\cos\theta-i\sin\theta.
Worked example(1+i)8(1+i)^8

Evaluate.

Goal
A single number.
Strategy
Modulus–argument form first.
  1. 1+i=2(cos⁡π4+isin⁡π4)1+i=\sqrt2\left(\cos\tfrac\pi4+i\sin\tfrac\pi4\right)
    ∣1+i∣=2|1+i|=\sqrt2, arg⁡=π4\arg=\frac\pi4.
  2. (1+i)8=(2)8(cos⁡2π+isin⁡2π)(1+i)^8=(\sqrt2)^8(\cos2\pi+i\sin2\pi)
    De Moivre.
  3. =16=16
    (2)8=24(\sqrt2)^8=2^4.
Answer1616.
✓ Check(1+i)2=2i(1+i)^2=2i, (2i)4=16(2i)^4=16 ✓.

2. Multiple-angle identities

cos⁡3θ\cos3\theta and sin⁡3θ\sin3\theta
  1. cos⁡3θ+isin⁡3θ=(c+is)3\cos3\theta+i\sin3\theta=(c+is)^3
    De Moivre with c=cos⁡θc=\cos\theta, s=sin⁡θs=\sin\theta.
  2. =c3+3c2(is)+3c(is)2+(is)3=c^3+3c^2(is)+3c(is)^2+(is)^3
    Binomial expansion.
  3. =(c3−3cs2)+i(3c2s−s3)=(c^3-3cs^2)+i(3c^2s-s^3)
    i2=−1i^2=-1, i3=−ii^3=-i.
  4. cos⁡3θ=c3−3c(1−c2)=4c3−3c\cos3\theta=c^3-3c(1-c^2)=4c^3-3c
    Real parts; s2=1−c2s^2=1-c^2.
  5. sin⁡3θ=3(1−s2)s−s3=3s−4s3\sin3\theta=3(1-s^2)s-s^3=3s-4s^3
    Imaginary parts.
Powers of cos⁡θ\cos\theta in multiple angles
  1. z=eiθ⇒z+z−1=2cos⁡θ, zn+z−n=2cos⁡nθz=e^{i\theta}\Rightarrow z+z^{-1}=2\cos\theta,\ z^n+z^{-n}=2\cos n\theta
    Add eiθe^{i\theta} and e−iθe^{-i\theta}.
  2. (2cos⁡θ)3=(z+z−1)3=z3+3z+3z−1+z−3(2\cos\theta)^3=(z+z^{-1})^3=z^3+3z+3z^{-1}+z^{-3}
    Binomial.
  3. 8cos⁡3θ=2cos⁡3θ+6cos⁡θ8\cos^3\theta=2\cos3\theta+6\cos\theta
    Pair the terms.
  4. cos⁡3θ=14(cos⁡3θ+3cos⁡θ)\cos^3\theta=\tfrac14(\cos3\theta+3\cos\theta)
    Ready to integrate.

3. Roots

Solving zn=reiαz^n=re^{i\alpha}
  1. reiα=rei(α+2kπ)re^{i\alpha}=re^{i(\alpha+2k\pi)}
    The argument is only defined up to whole turns.
  2. z=r1/nei(α+2kπ)/nz=r^{1/n}e^{i(\alpha+2k\pi)/n}
    Take the nnth root of each form.
  3. k=0,1,…,n−1k=0,1,\ldots,n-1
    k=nk=n repeats k=0k=0: exactly nn distinct roots.
The roots form a regular nn-gon centred at the origin, so they sum to 00 (for n≥2n\ge2).
Worked examplez3=8iz^3=8i

Find all roots in exponential form, −π<arg⁡z≤π-\pi<\arg z\le\pi.

Goal
Three roots.
Strategy
8i=8eiπ/28i=8e^{i\pi/2}.
  1. ∣z∣=81/3=2|z|=8^{1/3}=2
    Modulus.
  2. arg⁡z=π/2+2kπ3=π6, 5π6, 3π2\arg z=\frac{\pi/2+2k\pi}3=\frac\pi6,\ \frac{5\pi}6,\ \frac{3\pi}2
    k=0,1,2k=0,1,2.
  3. 3π2→−π2\frac{3\pi}2\to-\frac\pi2
    Bring into range.
Answer2eiπ/62e^{i\pi/6}, 2e5iπ/62e^{5i\pi/6}, 2e−iπ/2=−2i2e^{-i\pi/2}=-2i.
✓ Check(−2i)3=−8i3=8i(-2i)^3=-8i^3=8i ✓.
Common misconceptionOnly one root Dividing the argument by nn gives just one root. Always add 2kπ2k\pi first.
Key takeawayCube roots of unity 1,ω,ω21,\omega,\omega^2 with ω=e2πi/3\omega=e^{2\pi i/3} satisfy 1+ω+ω2=01+\omega+\omega^2=0 and ω3=1\omega^3=1.
Pause and think

What is the product of all nnth roots of unity?

Summary

  • Raise modulus, multiply argument.
  • Expand (c+is)n(c+is)^n for cos⁡nθ\cos n\theta, sin⁡nθ\sin n\theta.
  • zn+z−n=2cos⁡nθz^n+z^{-n}=2\cos n\theta for powers.
  • nn evenly spaced roots.
Key words
Root of unity
Solution of zn=1z^n=1.
Principal argument
Argument in (−π,π](-\pi,\pi].
HistoryAbraham de Moivre. A French mathematician living in London, de Moivre found the formula around 1707. He reputedly earned a living solving probability problems for gamblers in a coffee house.
Fun factRoots of unity in your music. The Fast Fourier Transform, used in every MP3 and phone call, works by evaluating polynomials at roots of unity.
Section 3 of 14

Key formulae

De Moivre's theorem[r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\left[r(\cos\theta+i\sin\theta)\right]^n=r^n(\cos n\theta+i\sin n\theta)
nth rootszn=reiα⇒z=r1/nei(α+2kπ)/n, k=0,1,…,n−1z^n=re^{i\alpha}\Rightarrow z=r^{1/n}e^{i(\alpha+2k\pi)/n},\ k=0,1,\ldots,n-1
Useful resultsz+1z=2cos⁡θ,zn+1zn=2cos⁡nθ(z=eiθ)z+\frac1z=2\cos\theta,\qquad z^n+\frac1{z^n}=2\cos n\theta\quad(z=e^{i\theta})
TrickRoots are evenly spaced. Find one root, then keep adding 2πn\frac{2\pi}n to the argument. The nn roots form a regular nn-gon centred on the origin, so they always sum to zero (for n≥2n\ge2).
Section 4 of 14

Worked examples

Example 1 · A power0 / 2 steps
Find (1+i)8(1+i)^8.
    Example 2 · cos 3θ0 / 3 steps
    Show cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta.
      Section 5 of 14

      Explore the graph

      The nth roots of unity
      Loading the interactive graph…
      Try this
      1. Step kk from 00 to n−1n-1: the root moves round the unit circle in equal steps of 2πn\frac{2\pi}n.
      2. At k=nk=n you're back at 11: there are exactly nn distinct roots.

      e2πik/ne^{2\pi ik/n} for k=0,…,n−1k=0,\ldots,n-1.

      In the real worldElectrical engineering. Three-phase power uses three voltages 120∘120^\circ apart: exactly the cube roots of unity. Because they add to zero, the return wire carries almost no current.
      Section 6 of 14

      Common mistakes

      IncorrectOnly one nnth root
      Correctnn distinct roots
      Add 2kπ2k\pi to the argument before dividing by nn.
      Incorrect(reiθ)n=reinθ(r e^{i\theta})^n=re^{in\theta}
      Correctrneinθr^ne^{in\theta}
      The modulus is raised to the power too.
      IncorrectArguments outside the required range
      CorrectAdjust by 2π2\pi
      Questions usually want −π<θ≤π-\pi<\theta\le\pi.

      Me expanding (cos θ + i sin θ)⁷ by hand

      De Moivre: just multiply the angle by 7.

      Section 7 of 14

      Exam tip

      Exam tipBinomial plus de Moivre. For "express cos⁡5θ\cos5\theta in powers of cos⁡θ\cos\theta", expand (c+is)5(c+is)^5 with Pascal's triangle and take the real part. Replace s2s^2 by 1−c21-c^2 at the end.
      Section 8 of 14

      Guided practice

      SA-FP-DM-G012 marksL1 FoundationNon-calculatorPowers
      Find the real number (cos⁡π6+isin⁡π6)6\left(\cos\frac\pi6+i\sin\frac\pi6\right)^6.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-DM-0012 marksL2 StandardNon-calculatorPowers
      Find (1+i)8(1+i)^8.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-DM-0022 marksL2 StandardNon-calculatorTrig identities
      cos⁡3θ=acos⁡3θ+bcos⁡θ\cos3\theta=a\cos^3\theta+b\cos\theta. Find aa.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-DM-0033 marksL3 ExamNon-calculatorRoots
      Solve z3=8iz^3=8i. Give the arguments of the roots in degrees, in the range −180<θ≤180-180<\theta\le180.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-DM-0041 markL2 StandardNon-calculatorRoots
      Find the sum of all the roots of z7=1z^7=1.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-DM-0053 marksL4 ChallengingNon-calculatorRoots
      ω\omega is a complex cube root of unity (ω≠1\omega\neq1). Find (1+ω)(1+ω2)(1+\omega)(1+\omega^2).

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Power: raise modulus, multiply argument.
      • Identities: expand and compare real/imaginary parts.
      • nn roots, evenly spaced, summing to zero.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-DM-Q011 markL1 FoundationNon-calculatorRoots
      How many distinct solutions does z6=−64z^6=-64 have?
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      9231 AS & A Level Further Mathematics
      Where
      Paper 2 · 2.4 Complex numbers
      Expected
      De Moivre's theorem for integer n; expressing cos nθ, sin nθ and powers of sin θ, cos θ; nth roots of unity and of general complex numbers; summing series.
      Calculator
      Permitted on all papers; exact answers are expected where the question asks.
      Typical style
      A typical question proves an identity for cos 5θ, then uses it to solve a polynomial equation exactly.
      Open in
      9231 · Paper 2
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

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      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).