{"id":1283,"date":"2015-03-25T21:19:32","date_gmt":"2015-03-25T20:19:32","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=1283"},"modified":"2019-05-04T08:56:07","modified_gmt":"2019-05-04T07:56:07","slug":"dm-guide-nombres-complexes-n2","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=1283","title":{"rendered":"DM guid\u00e9 Nombres complexes n\u00b02"},"content":{"rendered":"

Sujet :\u00a0Cliquer ici<\/a><\/strong>
\nCorrection en fin de cette page<\/p>\n

    \n
  1. Point invariant par f
    \n[peekaboo_link name=\u00a0\u00bbquestion1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″]Un point M est invariant si son affixe z v\u00e9rifie z ‘ = z[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion2″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]A\u00a0l’\u00e9quation z ‘ = z, on trouve comme unique solution z = i [\/peekaboo_content]<\/li>\n
  2. b. [peekaboo_link name=\u00a0\u00bbquestion3″]Aide pour les distances[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3″] Il faut passer aux\u00a0modules dans l’\u00e9galit\u00e9 du 2.a. et se rappeler qu’un module est une distance[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]R\u00e9ponse pour les distances[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb]MM’ = AM[\/peekaboo_content]
    \n<\/span>[peekaboo_link name=\u00a0\u00bbquestion4″]Aide\u00a0<\/span>pour les angles[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]\u00a0Il faut\u00a0passer aux arguments\u00a0dans l’\u00e9galit\u00e9 du 2.a. et se rappeler qu’un argument\u00a0est un angle[\/peekaboo_content]
    \n<\/span>[peekaboo_link name=\u00a0\u00bbquestion4b\u00a0\u00bb]R\u00e9ponse pour les angles[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4b\u00a0\u00bb] <\/span><\/span>\u00a0(\\overrightarrow{MA}, \\overrightarrow{MM}‘) \u00a0= – \\dfrac{\\pi}{2}[\/peekaboo_content]<\/span><\/li>\n
  3. Construction de B’
    \n[peekaboo_link name=\u00a0\u00bbquestion4c\u00a0\u00bb]Aide 1 [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4c\u00a0\u00bb]Utiliser les propri\u00e9t\u00e9s sur les distances et angles trouv\u00e9es en 2.b\u00a0[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion5″]Aide 2 [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion5″]Remplacer M par B et M ‘ par B ‘ dans les propri\u00e9t\u00e9s du 2.b[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion10″]Aide 3 [\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion10″]Les angles nous indiquent que les demi-droites\u00a0[BA) et [BB’) son perpendiculaires. Attention \u00e0 l’orientation de l’angle ![\/peekaboo_content]
    \n<\/span><\/li>\n
  4. b.
    \n[peekaboo_link name=\u00a0\u00bbquestion6″]Aide 1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion6″]Si M appartient au cercle de centre C, de rayon 2, alors\u00a0CM = 2, et\u00a0traduire cette information en terme de module[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion12″]Aide 2[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion12″]Pour trouver une information sur le point M ‘ , il faut passer au module dans l’\u00e9galit\u00e9 donn\u00e9e en 4.a. et utiliser le fait que le module de z – 2 est \u00e9gal \u00e0 2 si M appartient au cercle de centre C de rayon 2[\/peekaboo_content]
    \n<\/span> [peekaboo_link name=\u00a0\u00bbquestion13″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion13″] On trouve que M ‘ appartient au cercle de centre B de rayon 2 \\sqrt{2} [\/peekaboo_content]<\/li>\n<\/ol>\n

    Correction :\u00a0cliquer ici<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

    Sujet :\u00a0Cliquer ici Correction en fin de cette page Point invariant par f [peekaboo_link name=\u00a0\u00bbquestion1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″]Un point M est invariant si son affixe z v\u00e9rifie z ‘ = z[\/peekaboo_content] [peekaboo_link name=\u00a0\u00bbquestion2″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]A\u00a0l’\u00e9quation z ‘ = z, on trouve comme unique solution z = i [\/peekaboo_content] b. [peekaboo_link name=\u00a0\u00bbquestion3″]Aide pour les distances[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3″] Il faut passer […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1283"}],"collection":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1283"}],"version-history":[{"count":18,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1283\/revisions"}],"predecessor-version":[{"id":3420,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1283\/revisions\/3420"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1283"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1283"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}