Bonjour,
1)
z₁ = 2 + 2i
|z₁| = √(2² + 2²) = √(8) = 2√2
⇒ z₁ = 2√2(√2/2 + i√2/2) = 2√2(cos(π/4) + isin(π/4))
z₂ = -√3 - i
|z₂| = √(3 + 1) = 2
⇒ z₂ = 2(-√3/2 - i/2) = 2(cos(7π/6) + isin(7π/6))
z₃ = z₁/z₂
⇒ |z₃| = |z₁|/|z₂| = 2√2/2 = √2
et arg(z₃) = arg(z₁) - arg(z₂) = π/4 - 7π/6 = -11π/12
⇒ z₃ = √2(cos(-11π/12) + isin(-11π/12))
2) z₃ = z₁/z₂
= (2 + 2i)/(-√3 - i)
= (2 + 2i)(-√3 + i)/(-√3 - i)(-√3 + i)
= (-2√3 + 2i - 2i√3 - 2)/4
= (-√3 - 1)/2 + i(-√3 + 1)/2
3) on en déduit :
cos(13π/12) = cos(-11π/12) = (-√3 - 1)/2
et
sin(13π/12) = sin(-11π/12) = (-√3 + 1)/2
4)
sin(-13π/12) = - sin(13π/12) = (√3 - 1)/2
cos(π/12) = cos(13π/12 - π) = - cos(13π/12) = (√3 + 1)/2
5)
z = cos(13π/12) + isin(13π/12)
arg(z) = 13π/12
arg(zⁿ) = n x arg(z) = 13nπ/12
6) zⁿ ∈ R ⇒ arg(zⁿ) = 0 [π]
⇒ 13nπ/12 = kπ (k∈Z)
⇔ n = 12k/13
n est un entier ⇒ n minimal pour k = 13
Soit n = 12
