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De Moivre’s Theorem Applications

Key Terms

De Moivre’s theorem
For any integer n, (cosθ + i sinθ)n = cos(nθ) + i sin(nθ).
Powers of complex numbers
If z = r(cosθ + i sinθ), then zn = rn(cos(nθ) + i sin(nθ)).
Direction 1 — expressing trig of multiples
Expand (cosθ + i sinθ)n using the binomial theorem, equate real and imaginary parts to get cos(nθ) and sin(nθ) in terms of cosθ and sinθ.
Direction 2 — power reduction
Let z = cosθ + i sinθ. Then z + 1/z = 2cosθ and z − 1/z = 2i sinθ. Use (z + 1/z)n to express cosnθ as a sum of multiple angles.
Key substitution
zn + z−n = 2cos(nθ) and zn − z−n = 2i sin(nθ).
Summary of Key Results:
  • cos(2θ) = cos2θ − sin2θ = 2cos2θ − 1
  • sin(2θ) = 2sinθcosθ
  • cos(3θ) = 4cos3θ − 3cosθ
  • sin(3θ) = 3sinθ − 4sin3θ
  • cos2θ = ½(1 + cos(2θ))
  • cos4θ = ⅛(cos(4θ) + 4cos(2θ) + 3)
Hot Tip For Direction 1 (expanding cos(nθ)): expand using the binomial theorem, collect real parts (even powers of i give real terms). For Direction 2 (power reduction): expand (z + 1/z)n using the binomial theorem, pair terms like (zk + z−k) = 2cos(kθ). Always work with z = eiθ = cosθ + i sinθ.

Worked Example 1 — Express cos(3θ) in terms of cosθ

Consider (cosθ + i sinθ)3 = cos(3θ) + i sin(3θ) [De Moivre].

Expanding the LHS: cos3θ + 3cos2θ(i sinθ) + 3cosθ(i sinθ)2 + (i sinθ)3

= cos3θ + 3i cos2θsinθ − 3cosθsin2θ − i sin3θ

Real part: cos(3θ) = cos3θ − 3cosθsin2θ = cos3θ − 3cosθ(1−cos2θ) = 4cos3θ − 3cosθ.

Imaginary part: sin(3θ) = 3cos2θsinθ − sin3θ = 3(1−sin2θ)sinθ − sin3θ = 3sinθ − 4sin3θ.

Worked Example 2 — Express cos4θ as multiple angles

Let z = cosθ + i sinθ. Then z + z−1 = 2cosθ, so (2cosθ)4 = (z + z−1)4.

Expanding: z4 + 4z2 + 6 + 4z−2 + z−4 = (z4 + z−4) + 4(z2 + z−2) + 6

= 2cos(4θ) + 4×2cos(2θ) + 6 = 2cos(4θ) + 8cos(2θ) + 6.

So 16cos4θ = 2cos(4θ) + 8cos(2θ) + 6, giving cos4θ = ⅛[cos(4θ) + 4cos(2θ) + 3].

Proving De Moivre’s Theorem by Induction

De Moivre’s theorem states: for any positive integer n, (cosθ + i sinθ)n = cos(nθ) + i sin(nθ). The result also extends to negative integers and rational numbers, though the proof of the rational case requires additional care.

Proof by induction (positive integers):

Base case n=1: (cosθ + i sinθ)1 = cosθ + i sinθ = cos(1·θ) + i sin(1·θ). ✓

Inductive hypothesis: Assume (cosθ + i sinθ)k = cos(kθ) + i sin(kθ) for some k ≥ 1.

Inductive step: (cosθ + i sinθ)k+1 = (cosθ + i sinθ)k × (cosθ + i sinθ)

= [cos(kθ) + i sin(kθ)][cosθ + i sinθ]    [by inductive hypothesis]

= cos(kθ)cosθ − sin(kθ)sinθ + i[cos(kθ)sinθ + sin(kθ)cosθ]

= cos(kθ + θ) + i sin(kθ + θ) = cos((k+1)θ) + i sin((k+1)θ). ✓

Conclusion: By mathematical induction, De Moivre’s theorem holds for all positive integers n.

For n = 0: (cosθ + i sinθ)0 = 1 = cos(0) + i sin(0). For negative n: use z−n = (1/z)n and note that if z = cosθ + i sinθ, then 1/z = cosθ − i sinθ = cos(−θ) + i sin(−θ).

Direction 1: Expressing Trig Functions of Multiple Angles

Given De Moivre’s theorem, we can expand (cosθ + i sinθ)n via the binomial theorem and separate real and imaginary parts. This gives cos(nθ) as a polynomial in cosθ and sinθ.

For n = 4: The expansion gives 8 terms. Real parts (from even powers of i sinθ) give cos(4θ); imaginary parts give sin(4θ). After using sin2θ = 1 − cos2θ to eliminate sin, we express cos(4θ) purely in terms of cosθ:

cos(4θ) = 8cos4θ − 8cos2θ + 1

These are related to Chebyshev polynomials of the first kind. The imaginary part similarly gives sin(nθ)/sinθ as a polynomial in cosθ (for n ≥ 1), related to Chebyshev polynomials of the second kind.

Direction 2: Power Reduction Using z + 1/z

This direction is essential for integrating powers of trig functions. If z = eiθ = cosθ + i sinθ, then:

  • zn = einθ = cos(nθ) + i sin(nθ)
  • z−n = e−inθ = cos(nθ) − i sin(nθ)
  • zn + z−n = 2cos(nθ)
  • zn − z−n = 2i sin(nθ)

In particular: z + z−1 = 2cosθ, so (2cosθ)n = (z + z−1)n. Expand via binomial theorem, then pair terms symmetrically. Each pair zk + z−k = 2cos(kθ) reduces the power.

Similarly: z − z−1 = 2i sinθ, so (2i sinθ)n = (z − z−1)n. This allows power reduction for sinnθ.

Application: Integration of Powers of Trig Functions

The power-reduction identities derived from De Moivre’s theorem make integration tractable. For example:

∫ cos4θ dθ = ∫ ⅛[cos(4θ) + 4cos(2θ) + 3] dθ

= ⅛[sin(4θ)/4 + 2sin(2θ) + 3θ] + C

= sin(4θ)/32 + sin(2θ)/4 + 3θ/8 + C

Without De Moivre’s power reduction, this would require repeated use of the half-angle formula — a much longer process.

Computing Powers of Complex Numbers

De Moivre’s theorem makes computing large powers of complex numbers fast. Convert to polar form r(cosθ + i sinθ), apply zn = rn(cos(nθ) + i sin(nθ)), then convert back to Cartesian if needed.

Example: (1 + i)8. We have |1+i| = √2 and arg(1+i) = π/4, so 1+i = √2(cos(π/4) + i sin(π/4)).

Therefore (1+i)8 = (√2)8(cos(2π) + i sin(2π)) = 16(1 + 0i) = 16.

Exam Tip: Always convert to polar form before applying De Moivre’s theorem to a complex number power. Trying to expand directly (e.g. (1+i)8 via repeated multiplication) is extremely error-prone and wastes time.
Exam Tip: In Direction 1 proofs, always state explicitly which part (real or imaginary) you are equating. Do not just write “real part =” without expanding the LHS first.

Mastery Practice

  1. Use De Moivre’s theorem to find exact expressions. Fluency

    1. (a) Expand (cosθ + i sinθ)3 using the binomial theorem and hence show that cos(3θ) = 4cos3θ − 3cosθ.
    2. (b) Use the same expansion to show that sin(3θ) = 3sinθ − 4sin3θ.

  2. Compute the exact value. Fluency

    1. (a) Compute (1 + i)8 using De Moivre’s theorem. Express your answer as a real number.
    2. (b) Compute (√3 + i)10. Express your answer in the form a + bi.

  3. Express in terms of multiple angles. Fluency

    Let z = cosθ + i sinθ. Using z + z−1 = 2cosθ, expand (z + z−1)4 and hence show that cos4θ = ⅛[cos(4θ) + 4cos(2θ) + 3].

  4. Evaluate the integral using De Moivre’s power reduction. Fluency

    Using the result cos4θ = ⅛[cos(4θ) + 4cos(2θ) + 3], evaluate ∫0π/2 cos4θ dθ.

  5. Express sin5θ in terms of multiple angles. Understanding

    Let z = cosθ + i sinθ. Use z − z−1 = 2i sinθ to expand (z − z−1)5 and hence express sin5θ as a sum of sines of multiple angles. State the final result clearly.

  6. Prove a trigonometric identity. Understanding

    Use De Moivre’s theorem to show that:

    cos(5θ) = 16cos5θ − 20cos3θ + 5cosθ.

    Show all binomial expansion steps clearly.

  7. Integration application. Understanding

    (a) Use De Moivre’s theorem to express sin4θ in terms of multiple angles.
    (b) Hence evaluate ∫0π sin4θ dθ.

  8. Exact value from De Moivre’s. Understanding

    Using cos(3θ) = 4cos3θ − 3cosθ, substitute θ = π/9 to show that cos(π/3) = 4cos3(π/9) − 3cos(π/9) and hence show that cos(π/9) satisfies the cubic equation 8x3 − 6x − 1 = 0. Do not attempt to solve the cubic.

  9. Complex number application. Problem Solving

    Let z = 2(cos(π/5) + i sin(π/5)).

    1. (a) Find z5 and express as a Cartesian number.
    2. (b) Find z−3 and express in the form a + bi, leaving answers in exact form.
    3. (c) Hence find the exact value of |z5 − z−3|.

  10. Deriving and applying a power-sum identity. Problem Solving

    (a) Use z + z−1 = 2cosθ and z − z−1 = 2i sinθ to show that (2cosθ)2(2i sinθ)2 = (z2 − z−2)2 = −4sin2(2θ).

    (b) Hence show that cos2θsin2θ = ¼sin2(2θ) = ⅛(1 − cos(4θ)).

    (c) Use this result to evaluate ∫0π/2 cos2θ sin2θ dθ.

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