Quiz : Maîtrisez les nombres complexes en Terminale Scientifique

Question 1 sur 10 10:00

[{"id":47151,"question":"Quel est le module du nombre complexe z = 3 - 4i ?","option_a":"5","option_b":"7","option_c":"1","option_d":"√7","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le module de z = a + bi est donné par |z| = √(a² + b²). Ici, |z| = √(3² + (-4)²) = √(9 + 16) = √25 = 5.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47152,"question":"L'argument principal d'un nombre complexe est toujours compris entre 0 et π.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"L'argument principal est défini dans l'intervalle ]-π, π]. Par exemple, l'argument de -1 est π, mais celui de -i est -π\/2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47153,"question":"Soit z = 2(cos(π\/3) + i sin(π\/3)). Quelle est sa forme algébrique ?","option_a":"1 + i√3","option_b":"√3 + i","option_c":"2 + i2√3","option_d":"1 + i","option_e":"","option_f":"","bonne_reponse":"A","explication":"En développant z = 2(cos(π\/3) + i sin(π\/3)) = 2(1\/2 + i (√3\/2)) = 1 + i√3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47154,"question":"L'équation z² = -4 admet pour solutions :","option_a":"z = 2i ou z = -2i","option_b":"z = 2 ou z = -2","option_c":"z = 4i ou z = -4i","option_d":"Aucune solution réelle","option_e":"","option_f":"","bonne_reponse":"A","explication":"Les solutions de z² = -4 sont z = ±√(-4) = ±2i, car i² = -1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47155,"question":"Le conjugué d'un nombre complexe z = a + bi est toujours réel.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le conjugué de z = a + bi est z̄ = a - bi. Il est réel uniquement si b = 0 (cas où z est réel).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47156,"question":"Quelle est la forme trigonométrique de z = -1 - i ?","option_a":"√2 (cos(5π\/4) + i sin(5π\/4))","option_b":"√2 (cos(3π\/4) + i sin(3π\/4))","option_c":"2 (cos(π) + i sin(π))","option_d":"1 (cos(π\/4) + i sin(π\/4))","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le module de z est √((-1)² + (-1)²) = √2. L'argument est 5π\/4 (car z est dans le 3ème quadrant).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47157,"question":"Si z₁ = 2 + 3i et z₂ = 1 - i, alors z₁ + z₂ = ?","option_a":"3 + 2i","option_b":"1 + 4i","option_c":"3 - 2i","option_d":"2 + 2i","option_e":"","option_f":"","bonne_reponse":"A","explication":"L'addition de nombres complexes se fait composante par composante : (2+1) + (3-1)i = 3 + 2i.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47158,"question":"Pour tout nombre complexe z, on a |z|² = z × z̄.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"En effet, si z = a + bi, alors z̄ = a - bi. Donc z × z̄ = (a + bi)(a - bi) = a² + b² = |z|².","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47159,"question":"Résoudre l'équation z² - 2z + 5 = 0 dans ℂ.","option_a":"z = 1 + 2i ou z = 1 - 2i","option_b":"z = 2 + i ou z = 2 - i","option_c":"z = -1 + 2i ou z = -1 - 2i","option_d":"Pas de solution","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le discriminant Δ = (-2)² - 4×1×5 = 4 - 20 = -16. Les solutions sont z = (2 ± √(-16))\/2 = 1 ± 2i.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":47160,"question":"Le produit de deux nombres complexes de module 1 est toujours de module 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Si |z₁| = 1 et |z₂| = 1, alors |z₁ × z₂| = |z₁| × |z₂| = 1 × 1 = 1. Cette propriété est valable pour tout produit de complexes.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]