Quiz — Serie d_exercices Corrigés Complexes 1.pdf

Question 1 sur 10 20:00

[{"id":66461,"question":"Quel est le module du nombre complexe z = 3 - 4i ?","option_a":"5","option_b":"7","option_c":"25","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,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"5\", \"b\": \"7\", \"c\": \"25\", \"d\": \"√7\"}}","_debug_options_count":4},{"id":66462,"question":"L'argument principal d'un nombre complexe est toujours compris entre -π et π.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Par définition, l'argument principal d'un nombre complexe non nul est l'angle θ tel que -π \u003C θ ≤ π, mesuré dans le sens trigonométrique.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":66463,"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":"√3(cos(π\/3) + i sin(π\/3))","option_d":"2(cos(π\/4) + i sin(π\/4))","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le module est |z| = √(1² + (√3)²) = 2. L'argument θ vérifie tanθ = √3\/1 = √3, donc θ = π\/3. D'où z = 2(cos(π\/3) + i sin(π\/3)).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"2(cos(π\/3) + i sin(π\/3))\", \"b\": \"2(cos(π\/6) + i sin(π\/6))\", \"","_debug_options_count":4},{"id":66464,"question":"Le conjugué d'un nombre complexe z = a + bi est toujours égal à -z.","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, qui n'est égal à -z = -a - bi que si a = 0 (cas des nombres imaginaires purs).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":66465,"question":"Quelle est la solution de l'équation z² = -4 dans C ?","option_a":"z = 2i ou z = -2i","option_b":"z = 4i ou z = -4i","option_c":"z = 2 ou z = -2","option_d":"z = i ou z = -i","option_e":"","option_f":"","bonne_reponse":"a","explication":"z² = -4 ⇒ z = ±√(-4) = ±2i. Les solutions sont donc z = 2i et z = -2i.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"z = 2i ou z = -2i\", \"b\": \"z = 4i ou z = -4i\", \"c\": \"z = 2 ou z = ","_debug_options_count":4},{"id":66466,"question":"Pour tout nombre complexe z ≠ 0, |z * z̄| = |z|².","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"On a |z * z̄| = |z| * |z̄| = |z| * |z| = |z|², car |z̄| = |z|.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":66467,"question":"Soit z = 2e^(iπ\/4). Quelle est sa forme algébrique ?","option_a":"√2 + i√2","option_b":"1 + i√3","option_c":"2 + 2i","option_d":"√2 + i","option_e":"","option_f":"","bonne_reponse":"a","explication":"z = 2(cos(π\/4) + i sin(π\/4)) = 2(√2\/2 + i√2\/2) = √2 + i√2.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"√2 + i√2\", \"b\": \"1 + i√3\", \"c\": \"2 + 2i\", \"d\": \"√2 + i\"}}","_debug_options_count":4},{"id":66468,"question":"L'équation z³ = 8 admet trois solutions distinctes dans C.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"L'équation z³ = 8 a trois solutions dans C : z = 2, z = 2e^(2iπ\/3) et z = 2e^(4iπ\/3).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":66469,"question":"Quel est le résultat de (1 + i)^4 ?","option_a":"-4","option_b":"4i","option_c":"4","option_d":"-4i","option_e":"","option_f":"","bonne_reponse":"a","explication":"(1 + i)^2 = 1 + 2i + i² = 2i. Donc (1 + i)^4 = (2i)^2 = -4.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"-4\", \"b\": \"4i\", \"c\": \"4\", \"d\": \"-4i\"}}","_debug_options_count":4},{"id":66470,"question":"Pour tout nombre complexe z, Re(z) = (z + z̄)\/2.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Si z = a + bi, alors z̄ = a - bi. Donc (z + z̄)\/2 = (a + bi + a - bi)\/2 = 2a\/2 = a = Re(z).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]