Complex Number Question

Question

Solve for θ [such as in the polar form re^iθ] in (cos(-2pi/9) + isin(-2pi/9))^3.

Answer 1

(cos(-2π/9) + isin(-2π/9))³ = cos(- 2π/3) + isin(- 2π/3) = - 1/2 - i√3/2(Mouavr's Theorem)

Answer 2

Let z = cos(-2π/9) + i sin(-2π/9)

Since cosθ is an even function and sinθ is an odd function, cos(-θ) = cosθ and sin(-θ) = -sin(θ).

So, z = cos(2π/9) - i sin(2π/9)

r = magnitude of z = √ [cos²(2π/9) + sin²(2π/9)] = 1

By DeMoivre's Theorem,

z³ = r³ [ cos(3(2π/9)) - isin(3(2π/9)) ] = 1[ cos(2π/3) - isin(2π/3)] = -(1/2) - (√3/2) i