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Mayetyohann2
@Mayetyohann2
May 2019
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Bonsoir tout le monde svp.....Besoin d'aide!!! comment calculer le module et l'argument du nbre complexe Z=1+cosФ+isinФ (Ф): s'appel teta
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scoladan
Verified answer
Bonjour,
z = (1 + cosθ) + isinθ
|z| = √[(1 + cosθ)² + sin²θ]
= √[1 + 2cosθ + cos²θ + sin²θ]
= √(2 + 2cosθ)
= √[2(1 + cosθ)]
= √[2(2cos²(θ/2)]
= 2cos(θ/2)
⇒ z = |z|[(1 + cosθ)/|z| + isinθ/|z|]
⇔ z = 2cos(θ/2)[(2cos²(θ/2)/2cos(θ/2) + isinθ/2cos(θ/2)]
⇔ z = 2cos(θ/2)[cos(θ/2) + i2sin(θ/2)cos(θ/2)/2cos(θ/2)]
⇔ z = 2cos(θ/2)[cos(θ/2) + isin(θ/2)]
⇒ arg(z) = θ/2
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mayetyohann2
Mrci c'est la même rponse que j'ai trouvé.
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Verified answer
Bonjour,z = (1 + cosθ) + isinθ
|z| = √[(1 + cosθ)² + sin²θ]
= √[1 + 2cosθ + cos²θ + sin²θ]
= √(2 + 2cosθ)
= √[2(1 + cosθ)]
= √[2(2cos²(θ/2)]
= 2cos(θ/2)
⇒ z = |z|[(1 + cosθ)/|z| + isinθ/|z|]
⇔ z = 2cos(θ/2)[(2cos²(θ/2)/2cos(θ/2) + isinθ/2cos(θ/2)]
⇔ z = 2cos(θ/2)[cos(θ/2) + i2sin(θ/2)cos(θ/2)/2cos(θ/2)]
⇔ z = 2cos(θ/2)[cos(θ/2) + isin(θ/2)]
⇒ arg(z) = θ/2