Quiz : Maîtrisez les nombres complexes en Terminale Math
🧠 Quiz 10 questions 10 min
QUIZ INTERACTIFDiff. 5/10
Série complète d'exercices corrigés sur les nombres complexes pour les élèves de Terminale Math en Tunisie. Idéal pour réviser et préparer le baccalauréat.
Question 1 sur 10 10:00
[{"id":39975,"question":"Quel est le module du nombre complexe z = 3 - 4i ?","option_a":"5","option_b":"7","option_c":"√7","option_d":"√13","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le module est calculé par |z| = √(3² + (-4)²) = √(9 + 16) = √25 = 5.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39976,"question":"L'argument d'un nombre complexe non nul est toujours défini modulo 2π.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"L'argument θ d'un nombre complexe est défini à un multiple de 2π près, car les angles sont périodiques de période 2π.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39977,"question":"Soit z = 1 + i√3. Quelle est sa forme trigonométrique ?","option_a":"2(cos(π\/3) + i sin(π\/3))","option_b":"2(cos(π\/6) + i sin(π\/6))","option_c":"√2(cos(π\/4) + i sin(π\/4))","option_d":"2(cos(π\/3) - i sin(π\/3))","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le module est |z| = √(1² + (√3)²) = 2. L'argument θ vérifie cosθ = 1\/2 et sinθ = √3\/2, donc θ = π\/3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39978,"question":"Quelle est la solution de l'équation z² = -1 dans ℂ ?","option_a":"z = i","option_b":"z = 1","option_c":"z = -i","option_d":"z = i ou z = -i","option_e":"","option_f":"","bonne_reponse":"D","explication":"Les solutions sont z = i et z = -i, car i² = -1 et (-i)² = -1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39979,"question":"Le conjugué d'un nombre complexe z = a + bi est toujours un nombre 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 seulement si b = 0 (cas où z est réel).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39980,"question":"Quel est le résultat de (1 + i)³ ?","option_a":"2 + 2i","option_b":"-2 + 2i","option_c":"2 - 2i","option_d":"-2 - 2i","option_e":"","option_f":"","bonne_reponse":"B","explication":"En développant : (1 + i)³ = 1 + 3i + 3i² + i³ = 1 + 3i - 3 - i = -2 + 2i.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39981,"question":"Si z = 2e^(iπ\/4), quel est son argument principal ?","option_a":"π\/4","option_b":"π\/2","option_c":"3π\/4","option_d":"π","option_e":"","option_f":"","bonne_reponse":"A","explication":"La forme exponentielle z = re^(iθ) montre que l'argument principal est θ = π\/4.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39982,"question":"L'équation z² + 4z + 5 = 0 admet deux solutions réelles.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le discriminant est Δ = 16 - 20 = -4 \u003C 0. Les solutions sont donc complexes : z = -2 ± i.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39983,"question":"Quel est le produit de z₁ = 2(cos(π\/6) + i sin(π\/6)) et z₂ = 3(cos(π\/3) + i sin(π\/3)) ?","option_a":"6(cos(π\/2) + i sin(π\/2))","option_b":"5(cos(π\/6) + i sin(π\/6))","option_c":"6(cos(π\/6) + i sin(π\/6))","option_d":"5(cos(π\/2) + i sin(π\/2))","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le produit de deux nombres complexes en forme trigonométrique a pour module le produit des modules (2×3=6) et pour argument la somme des arguments (π\/6 + π\/3 = π\/2).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":39984,"question":"La forme algébrique de e^(iπ) est -1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"D'après la formule d'Euler, e^(iπ) = cos(π) + i sin(π) = -1 + i·0 = -1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]
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