De Moivre's Theorem

[MathDefinition title="De Moivre's Theorem"] For any complex number z = r(cos θ + i sin θ) and any integer n: zⁿ = rⁿ(cos nθ + i sin nθ) In exponential form: (re^(iθ))ⁿ = rⁿe^(inθ) [/MathDefinition]

[MathDefinition title="nth Roots of a Complex Number"] The nth roots of z = r(cos θ + i sin θ) are given by: z_k = r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, 2, ..., n−1 [/MathDefinition]

[MathKeyFormula title="Key Applications of De Moivre's Theorem"]

1. Finding powers: zⁿ = rⁿ(cos nθ + i sin nθ)
2. Finding roots: z^(1/n) = r^(1/n) cis((θ + 2kπ)/n)
3. Euler's formula: e^(iθ) = cos θ + i sin θ
4. Trigonometric identities:
   • cos nθ = Re[(cos θ + i sin θ)ⁿ]
   • sin nθ = Im[(cos θ + i sin θ)ⁿ]
5. (cos θ + i sin θ)⁻ⁿ = cos nθ − i sin nθ [/MathKeyFormula]

[MathNote title="Using De Moivre's Theorem"]

1. Always convert to polar form first
2. Apply the theorem by raising r to the power and multiplying θ by n
3. For roots, add 2kπ to θ before dividing by n
4. There are exactly n distinct nth roots [/MathNote]

[MathExample title="Example 1: Finding Power"] Problem: Find (1 + i)⁶ using De Moivre's theorem.

Solution: First, convert 1 + i to polar form: r = √(1² + 1²) = √2 θ = tan⁻¹(1/1) = π/4

So, 1 + i = √2 cis(π/4)

Using De Moivre's theorem: (1 + i)⁶ = (√2)⁶ cis(6 × π/4) = 2³ cis(3π/2) = 8[cos(3π/2) + i sin(3π/2)] = 8[0 + i(−1)] = −8i

Therefore, (1 + i)⁶ = −8i [/MathExample]

[MathExample title="Example 2: Finding Cube Root"] Problem: Find all cube roots of 8.

Solution: 8 = 8(cos 0 + i sin 0) = 8 cis(0)

Using the formula for nth roots: z_k = 8^(1/3) cis((0 + 2kπ)/3) for k = 0, 1, 2

z₀ = 2 cis(0) = 2(cos 0 + i sin 0) = 2 z₁ = 2 cis(2π/3) = 2(cos 120° + i sin 120°) = 2(−1/2 + i√3/2) = −1 + i√3 z₂ = 2 cis(4π/3) = 2(cos 240° + i sin 240°) = 2(−1/2 − i√3/2) = −1 − i√3

The cube roots of 8 are: 2, −1 + i√3, −1 − i√3 [/MathExample]

[MathExample title="Example 3: Proving Identity"] Problem: Prove that cos 3θ = 4cos³θ − 3cosθ using De Moivre's theorem.

Solution: Using De Moivre's theorem: (cos θ + i sin θ)³ = cos 3θ + i sin 3θ

Expanding the left side using binomial theorem: (cos θ + i sin θ)³ = cos³θ + 3cos²θ(i sin θ) + 3cosθ(i sin θ)² + (i sin θ)³ = cos³θ + 3i cos²θ sin θ − 3cosθ sin²θ − i sin³θ = (cos³θ − 3cosθ sin²θ) + i(3cos²θ sin θ − sin³θ)

Equating real parts: cos 3θ = cos³θ − 3cosθ sin²θ = cos³θ − 3cosθ(1 − cos²θ) = cos³θ − 3cosθ + 3cos³θ = 4cos³θ − 3cosθ

Hence proved. [/MathExample]

[MathExample title="Example 4: Expressing Trigonometric Powers"] Problem: Express cos⁴θ in terms of multiples of θ.

Solution: Using cos θ = (e^(iθ) + e^(−iθ))/2:

cos⁴θ = [(e^(iθ) + e^(−iθ))/2]⁴ = (1/16)(e^(iθ) + e^(−iθ))⁴

Using binomial expansion: = (1/16)[e^(4iθ) + 4e^(2iθ) + 6 + 4e^(−2iθ) + e^(−4iθ)] = (1/16)[(e^(4iθ) + e^(−4iθ)) + 4(e^(2iθ) + e^(−2iθ)) + 6] = (1/16)[2cos 4θ + 4(2cos 2θ) + 6] = (1/16)[2cos 4θ + 8cos 2θ + 6] = (1/8)cos 4θ + (1/2)cos 2θ + 3/8

Therefore, cos⁴θ = (3/8) + (1/2)cos 2θ + (1/8)cos 4θ [/MathExample]