Find the value of A and B......if w is the complex cube root of unity

Question

(1+w)⁷ = A+Bw

Answer 1

If w represents the complex cube root of unity then

w₁ = 1

w₂ = e^2πi/3

w₃ = e^4πi/3

We see that w₁ * w₂ = w₂, w₁ * w₃ = w₃, w₂ * w₂ = w₃, w₂ * w₃ = w₁

This is due to the "circularity" of the solutions...

If these solutions we can choose A,B from this choice...

(1 + e^2πi/3)⁷ = A + Be^2πi/3

We make the replacement of 1 --> w₁

(1 + e^2πi/3)⁷ = (w₁ + w₂)(w₁ + w₂)(w₁ + w₂)(w₁ + w₂)(w₁ + w₂)(w₁ + w₂)(w₁ + w₂)

= (w₁ + w₂)(w₁*w₁ + 2w₁*w₂ + w₂*w₂)³

= (w₁ + w₂)(w₁ + 2w₂ + w₃)³

= (w₁ + w₂)(w₁ + 2w₂ + w₃)(w₁ + 2w₂ + w₃)(w₁ + 2w₂ + w₃)

= (w₁ + w₂)(w₁ + 2w₂ + w₃)(w₁*w₁ + 4w₁*w₂ + 2w₁*w₃ + 4w₂*w₂ + 4w₂*w₃ + w₃*w₃)

= (w₁ + w₂)(w₁ + 2w₂ + w₃)(5 + 5w₂ + 6w₃)

= (w₁*w₁ + 3w₁*w₂ + w₁*w₃ +2w₂*w₂ +w₂*w₃)(5 + 5w₂ + 6w₃)

= (2 + 3w₂ + 3w₃)(5 + 5w₂ + 6w₃)

= (10 + 10w₂ + 12w₃ + 15w₂ + 15w₂*w₂ + 18w₂*w₃ + 15w₃ + 15w₂*w₃ + 18w₃*w₃)

= (43 + 43w₂ + 42w₃)

Therefore --> ( 1 + w₂ )⁷ = (43 + 43w₂ + 42w₃)

43 + 43w₂ + 42w₃ = A + Bw₂

Doing the small for w₃ yields -->

(1 + e^4πi/3)⁷ = (w₁ + w₃)(w₁ + w₃)(w₁ + w₃)(w₁ + w₃)(w₁ + w₃)(w₁ + w₃)(w₁ + w₃)

= (w₁ + w₃)(w₁ * w₁ + 2w₁ * w₃ + w₃ * w₃)³

= (w₁ + w₃)(w₁ + 2w₃ + w₂)³

= (w₁ + w₃)(w₁ + 2w₃ + w₂)(w₁ + 2w₃ + w₂)(w₁ + 2w₃ + w₂)

= (w₁ + w₃)(w₁ + 2w₃ + w₂)(w₁*w₁ + 4w₁*w₃ + 2w₁*w₂ + 4w₃*w₃ + 4w₂*w₃ + w₂*w₂)

= (w₁ + w₃)(w₁ + 2w₃ + w₂)(5 + 5w₃ + 6w₂)

= (w₁*w₁ + 3w₁*w₃ + w₁*w₂ + 2w₃*w₃ + w₂*w₃)(5 + 5w₃ + 6w₂)

= (2 + 3w₃ + 3w₂)(5 + 5w₃ + 6w₂)

= (10 + 10w₃ + 12w₂ + 15w₃ + 15w₃*w₃ + 18w₂*w₃ + 15w₂ + 15w₂*w₃ + 18w₂*w₂)

= (43 + 43w₃ + 42w₂)

Therefore --> ( 1 + w₃ )⁷ = (43 + 43w₃ + 42w₂)

43 + 43w₃ + 42w₂ = A + Bw₃

We have two equations in two unknowns...

43 + 43w₂ + 42w₃ = A + Bw₂

43 + 43w₃ + 42w₂ = A + Bw₃

Subtracting both equations yields --> -w₃ + w₂ = B(w₂ - w₃)

We gives us B = 1

With B = 1 we solve for A

43 + 43w₂ + 42w₃ = A + w2

A = 43 + 42(w₂ + w₃)

Therefore the solution gives B as a real number and A as a complex number