Question

PLEASE I NEED THIS NOW!! Please

Answer 1

Suppose that r (cos(θ) + i sin(θ)) is a third root of 27 (cos(π/5) + i sin(π/5)), i.e.

[r (cos(θ) + i sin(θ))]³ = 27 (cos(π/5) + i sin(π/5))

Expand the left side using de Moivre's theorem:

r³ (cos(3θ) + i sin(3θ)) = 27 (cos(π/5) + i sin(π/5))

Matching up real and imaginary parts, we have

r³ cos(3θ) = 27 cos(π/5)

r³ sin(3θ) = 27 sin(π/5)

Right away we see that r³ = 27 ⇒ r = 3, which eliminates A.

We also have

(r³ sin(3θ)) / (r³ cos(3θ)) = (27 sin(π/5)) / (27 cos(π/5))

tan(3θ) = tan(π/5)

3θ = tan⁻¹(tan(π/5))

3θ = π/5 + nπ

(where n is an integer)

θ = π/15 + nπ/3

Now,

• n = 0   ⇒   θ = π/15   ⇒   D is a third root

• n = 1   ⇒   θ = π/15 + π/3 = 2π/5

• n = 2   ⇒   θ = π/15 + 2π/3 = 11π/15   ⇒   C is a third root