Solutions — De Moivre’s Theorem Applications

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

(a) Show cos(3θ) = 4cos³θ − 3cosθ:

By De Moivre’s theorem: (cosθ + i sinθ)³ = cos(3θ) + i sin(3θ).

Expand the LHS using the binomial theorem:

(cosθ)³ + 3(cosθ)²(i sinθ) + 3(cosθ)(i sinθ)² + (i sinθ)³

= cos³θ + 3i cos²θ sinθ + 3cosθ(i²sin²θ) + i³sin³θ

= cos³θ + 3i cos²θ sinθ − 3cosθ sin²θ − i sin³θ

Equating real parts: cos(3θ) = cos³θ − 3cosθ sin²θ

= cos³θ − 3cosθ(1 − cos²θ) [using sin²θ = 1 − cos²θ]

= cos³θ − 3cosθ + 3cos³θ

cos(3θ) = 4cos³θ − 3cosθ ✓

(b) Show sin(3θ) = 3sinθ − 4sin³θ:

From the same expansion, equating imaginary parts:

sin(3θ) = 3cos²θ sinθ − sin³θ

= 3(1 − sin²θ)sinθ − sin³θ [using cos²θ = 1 − sin²θ]

= 3sinθ − 3sin³θ − sin³θ

sin(3θ) = 3sinθ − 4sin³θ ✓

Compute the exact value. Fluency

(a) Compute (1 + i)⁸:

Convert to polar form: |1+i| = √(1²+1²) = √2, arg(1+i) = π/4.

So 1+i = √2(cos(π/4) + i sin(π/4)).

By De Moivre: (1+i)⁸ = (√2)⁸(cos(8π/4) + i sin(8π/4))

= 2⁴(cos(2π) + i sin(2π)) = 16(1 + 0i) = 16

(b) Compute (√3 + i)¹⁰:

|√3 + i| = √(3+1) = 2, arg(√3+i) = π/6 (since tanθ = 1/√3).

So √3+i = 2(cos(π/6) + i sin(π/6)).

By De Moivre: (√3+i)¹⁰ = 2¹⁰(cos(10π/6) + i sin(10π/6))

= 1024(cos(5π/3) + i sin(5π/3))

cos(5π/3) = 1/2, sin(5π/3) = −√3/2

= 1024(1/2 − i√3/2) = 512 − 512√3 i

Express in terms of multiple angles. Fluency

Let z = cosθ + i sinθ, so z⁻¹ = cosθ − i sinθ and z + z⁻¹ = 2cosθ.

Therefore (2cosθ)⁴ = (z + z⁻¹)⁴.

Expand using the binomial theorem:

(z + z⁻¹)⁴ = z⁴ + 4z³·z⁻¹ + 6z²·z⁻² + 4z·z⁻³ + z⁻⁴

= z⁴ + 4z² + 6 + 4z⁻² + z⁻⁴

= (z⁴ + z⁻⁴) + 4(z² + z⁻²) + 6

Using zⁿ + z⁻ⁿ = 2cos(nθ):

16cos⁴θ = 2cos(4θ) + 4 × 2cos(2θ) + 6 = 2cos(4θ) + 8cos(2θ) + 6

Dividing by 16: cos⁴θ = ⅛[cos(4θ) + 4cos(2θ) + 3] ✓

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

Using cos⁴θ = ⅛[cos(4θ) + 4cos(2θ) + 3]:

∫₀^π/2 cos⁴θ dθ = ⅛ ∫₀^π/2 [cos(4θ) + 4cos(2θ) + 3] dθ

= ⅛ [sin(4θ)/4 + 2sin(2θ) + 3θ]₀^π/2

At θ = π/2: sin(2π)/4 + 2sin(π) + 3π/2 = 0 + 0 + 3π/2 = 3π/2

At θ = 0: 0 + 0 + 0 = 0

= ⅛ × 3π/2 = 3π/16

Express sin⁵θ in terms of multiple angles. Understanding

Let z = cosθ + i sinθ. Then z − z⁻¹ = 2i sinθ, so (2i sinθ)⁵ = (z − z⁻¹)⁵.

Expand the RHS using the binomial theorem:

(z − z⁻¹)⁵ = z⁵ − 5z³ + 10z − 10z⁻¹ + 5z⁻³ − z⁻⁵

Group symmetric pairs:

= (z⁵ − z⁻⁵) − 5(z³ − z⁻³) + 10(z − z⁻¹)

Using zⁿ − z⁻ⁿ = 2i sin(nθ):

= 2i sin(5θ) − 5 × 2i sin(3θ) + 10 × 2i sin(θ)

= 2i[sin(5θ) − 5sin(3θ) + 10sin(θ)]

The LHS is (2i)⁵sin⁵θ = 32i⁵sin⁵θ = 32i sin⁵θ [since i⁵ = i]

So: 32i sin⁵θ = 2i[sin(5θ) − 5sin(3θ) + 10sin(θ)]

Dividing by 2i: 16sin⁵θ = sin(5θ) − 5sin(3θ) + 10sin(θ)

sin⁵θ = &frac{1}{16}[sin(5θ) − 5sin(3θ) + 10sin(θ)]

Prove a trigonometric identity. Understanding

By De Moivre: (cosθ + i sinθ)⁵ = cos(5θ) + i sin(5θ).

Expand LHS using the binomial theorem (C(5,k) = 1, 5, 10, 10, 5, 1):

= cos⁵θ + 5cos⁴θ(i sinθ) + 10cos³θ(i sinθ)² + 10cos²θ(i sinθ)³ + 5cosθ(i sinθ)⁴ + (i sinθ)⁵

= cos⁵θ + 5i cos⁴θsinθ − 10cos³θsin²θ − 10i cos²θsin³θ + 5cosθsin⁴θ + i sin⁵θ

Equating real parts:

cos(5θ) = cos⁵θ − 10cos³θsin²θ + 5cosθsin⁴θ

Substitute sin²θ = 1 − cos²θ and sin⁴θ = (1−cos²θ)² = 1 − 2cos²θ + cos⁴θ:

= cos⁵θ − 10cos³θ(1−cos²θ) + 5cosθ(1−2cos²θ+cos⁴θ)

= cos⁵θ − 10cos³θ + 10cos⁵θ + 5cosθ − 10cos³θ + 5cos⁵θ

= (1+10+5)cos⁵θ + (−10−10)cos³θ + 5cosθ

cos(5θ) = 16cos⁵θ − 20cos³θ + 5cosθ ✓

Integration application. Understanding

(a) Express sin⁴θ in terms of multiple angles:

z − z⁻¹ = 2i sinθ, so (2i sinθ)⁴ = (z − z⁻¹)⁴.

Expand: z⁴ − 4z² + 6 − 4z⁻² + z⁻⁴

= (z⁴ + z⁻⁴) − 4(z² + z⁻²) + 6

= 2cos(4θ) − 8cos(2θ) + 6

LHS: (2i)⁴sin⁴θ = 16i⁴sin⁴θ = 16sin⁴θ

So 16sin⁴θ = 2cos(4θ) − 8cos(2θ) + 6

sin⁴θ = ⅛[cos(4θ) − 4cos(2θ) + 3]

(b) Evaluate ∫₀^π sin⁴θ dθ:

= ⅛ ∫₀^π [cos(4θ) − 4cos(2θ) + 3] dθ

= ⅛ [sin(4θ)/4 − 2sin(2θ) + 3θ]₀^π

At θ = π: sin(4π)/4 − 2sin(2π) + 3π = 0 − 0 + 3π = 3π

At θ = 0: 0

= ⅛ × 3π = 3π/8

Exact value from De Moivre’s. Understanding

From Q1(a): cos(3θ) = 4cos³θ − 3cosθ.

Substitute θ = π/9:

cos(3 × π/9) = 4cos³(π/9) − 3cos(π/9)

cos(π/3) = 4cos³(π/9) − 3cos(π/9) ✓

Let x = cos(π/9). Since cos(π/3) = 1/2:

4x³ − 3x = 1/2

Multiply through by 2: 8x³ − 6x = 1

8x³ − 6x − 1 = 0

Therefore cos(π/9) satisfies the cubic equation 8x³ − 6x − 1 = 0. ✓

Complex number application. Problem Solving

z = 2(cos(π/5) + i sin(π/5)), so r = 2 and θ = π/5.

(a) Find z⁵:

z⁵ = 2⁵(cos(5π/5) + i sin(5π/5)) = 32(cos(π) + i sin(π)) = 32(−1 + 0i) = −32

(b) Find z⁻³:

z⁻³ = 2⁻³(cos(−3π/5) + i sin(−3π/5)) = (1/8)(cos(3π/5) − i sin(3π/5))

cos(3π/5) = cos(108°) = −cos(72°) = −(√5−1)/4

sin(3π/5) = sin(108°) = sin(72°) = √(10+2√5)/4

z⁻³ = (1/8)[−(√5−1)/4 − i√(10+2√5)/4]

= −(√5−1)/32 − i√(10+2√5)/32

(c) Find |z⁵ − z⁻³|:

z⁵ = −32 (purely real)

z⁻³ has modulus |z⁻³| = 2⁻³ = 1/8

z⁵ − z⁻³ has real part: −32 − (−(√5−1)/32) = −32 + (√5−1)/32 = (−1024 + √5 − 1)/32 = (−1025 + √5)/32

imaginary part: −(−√(10+2√5)/32) = √(10+2√5)/32

|z⁵ − z⁻³| = √[((−1025+√5)/32)² + (√(10+2√5)/32)²]

Since |z⁵| = 32 and |z⁻³| = 1/8, and their arguments differ by π − (−3π/5) = 8π/5:

|z⁵ − z⁻³|² = |z⁵|² + |z⁻³|² − 2|z⁵||z⁻³|cos(8π/5)

= 1024 + 1/64 − 2(32)(1/8)cos(8π/5)

cos(8π/5) = cos(2π − 2π/5) = cos(2π/5) = (√5−1)/4

= 1024 + 1/64 − 8(√5−1)/4 = 1024 + 1/64 − 2(√5−1)

= 1024 + 1/64 − 2√5 + 2 = 1026 + 1/64 − 2√5

|z⁵ − z⁻³| = √(1026 + 1/64 − 2√5)

Note: ≈ √(1026 + 0.0156 − 4.472) ≈ √1021.5 ≈ 31.96

Deriving and applying a power-sum identity. Problem Solving

(a) Show (2cosθ)²(2i sinθ)² = −4sin²(2θ):

(2cosθ)² = (z + z⁻¹)² and (2i sinθ)² = (z − z⁻¹)²

Product: (z + z⁻¹)²(z − z⁻¹)² = [(z + z⁻¹)(z − z⁻¹)]²

= [z² − z⁻²]²

Now z² − z⁻² = 2i sin(2θ), so [z² − z⁻²]² = [2i sin(2θ)]² = 4i²sin²(2θ) = −4sin²(2θ) ✓

(b) Hence show cos²θsin²θ = ⅛(1 − cos(4θ)):

From (a): (2cosθ)²(2i sinθ)² = −4sin²(2θ)

LHS = 4cos²θ × −4sin²θ = −16cos²θsin²θ

So −16cos²θsin²θ = −4sin²(2θ)

cos²θsin²θ = &frac{1}{4}sin²(2θ) ✓

Using sin²(2θ) = ½(1 − cos(4θ)):

cos²θsin²θ = &frac{1}{4} × ½(1 − cos(4θ)) = ⅛(1 − cos(4θ)) ✓

(c) Evaluate ∫₀^π/2 cos²θ sin²θ dθ:

= ∫₀^π/2 ⅛(1 − cos(4θ)) dθ

= ⅛ [θ − sin(4θ)/4]₀^π/2

At θ = π/2: π/2 − sin(2π)/4 = π/2 − 0 = π/2