Les secrets du nombre j : algèbre et trigonométrie en Terminale

Question 1 sur 10 10:00

[{"id":13483,"question":"Quelle est la valeur de j³ ?","option_a":"j","option_b":"1","option_c":"-1","option_d":"j²","option_e":"","option_f":"","bonne_reponse":"B","explication":"Par définition, j est une racine cubique de l'unité, donc j³ = 1.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13484,"question":"L'égalité 1 + j + j² = 0 est-elle vraie ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C'est une propriété fondamentale des racines cubiques de l'unité : la somme des trois racines est nulle.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13485,"question":"Que vaut j² en fonction de j ?","option_a":"j","option_b":"-1-j","option_c":"1","option_d":"j+1","option_e":"","option_f":"","bonne_reponse":"B","explication":"De 1 + j + j² = 0, on déduit j² = -1 - j.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13486,"question":"Quel est le module de j ?","option_a":"0","option_b":"1","option_c":"√3\/2","option_d":"2","option_e":"","option_f":"","bonne_reponse":"B","explication":"j est une racine de l'unité, donc son module est 1 (|j| = 1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13487,"question":"L'équation x³ - 1 = 0 admet-elle j comme solution ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"j est bien une racine cubique de 1, donc solution de x³ - 1 = 0.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13488,"question":"Que vaut (1 + j)² ?","option_a":"1 + 2j","option_b":"j","option_c":"-j","option_d":"1","option_e":"","option_f":"","bonne_reponse":"C","explication":"(1 + j)² = 1 + 2j + j² = 1 + 2j -1 -j = j.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13489,"question":"La forme trigonométrique de j est :","option_a":"cos(2π\/3) + i sin(2π\/3)","option_b":"cos(π\/3) + i sin(π\/3)","option_c":"cos(π\/2) + i sin(π\/2)","option_d":"cos(π) + i sin(π)","option_e":"","option_f":"","bonne_reponse":"A","explication":"j = e^(2iπ\/3) = cos(2π\/3) + i sin(2π\/3).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13490,"question":"L'expression j⁴ + j⁵ est égale à :","option_a":"0","option_b":"1","option_c":"j","option_d":"-1","option_e":"","option_f":"","bonne_reponse":"B","explication":"j⁴ = j et j⁵ = j², donc j⁴ + j⁵ = j + j² = -1 (car 1 + j + j² = 0).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13491,"question":"Peut-on écrire j sous forme algébrique a + bi avec a et b réels ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Oui, j = -1\/2 + i√3\/2, donc a = -1\/2 et b = √3\/2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13492,"question":"Quel est le résultat de (1 - j)(1 - j²) ?","option_a":"1","option_b":"3","option_c":"0","option_d":"2","option_e":"","option_f":"","bonne_reponse":"B","explication":"(1 - j)(1 - j²) = 1 - j - j² + j³ = 1 - j - j² + 1 = 2 - (j + j²) = 2 - (-1) = 3 (car j + j² = -1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]