{"id":234650,"date":"2022-06-13T17:14:51","date_gmt":"2022-06-13T15:14:51","guid":{"rendered":"https:\/\/sherpas.com\/blog\/?p=234650"},"modified":"2025-09-29T11:28:23","modified_gmt":"2025-09-29T09:28:23","slug":"argument-dun-nombre-complexe","status":"publish","type":"post","link":"https:\/\/sherpas.com\/blog\/argument-dun-nombre-complexe\/","title":{"rendered":"Comment d\u00e9terminer l’argument d’un nombre complexe ?"},"content":{"rendered":"\n

Vous \u00e9tudiez actuellement les notions li\u00e9es \u00e0 l\u2019argument des nombres complexes <\/strong>? Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9, ma\u00eetrisez avec aisance toutes les subtilit\u00e9s de ce chapitre, gr\u00e2ce \u00e0 des m\u00e9thodologies compl\u00e8tes et adapt\u00e9es. <\/p>\n\n\n\n

En compl\u00e9ment, nos cours en ligne de maths<\/a><\/strong> te permettront d’int\u00e9grer les subtilit\u00e9s de la forme trigonom\u00e9trique. \ud83d\udcbc <\/p>\n\n\n\n

Argument d’un nombre complexe<\/h2>\n\n\n\n

D\u00e9finition<\/h3>\n\n\n\n

<\/p>\nSoit \"z\" un nombre complexe non nul. L’ensemble des arguments de \"z\" est l’ensemble des ant\u00e9c\u00e9dents de \"\dfrac{z}{|z|}\" par l’application exponentielle complexe d\u00e9finie dans la partie pr\u00e9c\u00e9dente.
\nUn argument de \"z\" sera not\u00e9 \"\arg(z)\".\n\n\n\n

Remarque<\/h4>\n\n\n\n

<\/p>\nSi \"\theta_0\" est un argument de \"z\", l’ensemble des arguments de \"z\" est donn\u00e9 par \"\big\{\theta_0+2k\pi,\;k\in\Z\big\}\". On dit que les arguments de \"z\" sont tous \u00e9gaux modulo \"2\pi\", on note \"\arg(z)\equiv \theta_0\,[2\pi]\". En g\u00e9n\u00e9ral, on choisira un argument contenu dans \"[0, 2\pi[\" ou dans \"] - \pi, \pi]\".\n\n\n\n

Exemples<\/h4>\n\n\n\n

<\/p>\n

  • \"\arg(1)=0\,[2\pi]\" <\/li>\n
  • \"\arg(\i)=0\,[\dfrac{\pi}{2}]\" <\/li>\n
  • \"\arg(-1)=\pi\,[2\pi]\" <\/li>\n\n\n\n

    Proposition<\/h3>\n\n\n\n

    <\/p>\nSoit \"z\" un nombre complexe non nul et \"\theta\" un de ses arguments. Alors :\n

      <\/span>   <\/span>\"\[z=|z|e^{\i\theta}.\]\"<\/p>\nCette expression est appel\u00e9e forme trigonom\u00e9trique de \"z\".\n\n\n\n

    Remarque<\/h4>\n\n\n\n

    <\/p>\nOn rencontre souvent la notation \"\theta \in ] - \pi, \pi]\" pour l’argument de \"z\in\mathbb{C}^*\" et \"\rho\in\mathbb{R}_+^*\" pour son module. La forme trigonom\u00e9trique de \"z\" s’\u00e9crit alors \"z=\rho e^{\i \theta}\".\n\n\n\n

    Proposition<\/h3>\n\n\n\n

    <\/p>\nSoient deux complexes \"z\" et \"z'\" avec \"z' \neq 0\" tels que : \"z=\rho e^{\i\theta}\" et \"z'=\rho 'e^{\i\theta'}\". Alors :\n

  • \"zz'=\rho\rho 'e^{\i(\theta+\theta')}\". <\/li>\n
  • \"\dfrac{z}{z'}=\dfrac{\rho}{\rho '}e^{\i(\theta-\theta')}\". <\/li>\n\n\n\n

    D\u00e9monstration<\/h3>\n\n\n\n

    Ces r\u00e9sultats se d\u00e9duisent directement des propri\u00e9t\u00e9s de l\u2019exponentielle complexe.<\/p>\n\n\n\n

    Corollaire<\/h3>\n\n\n\n

    <\/p>\nSoient \"z\" et \"z'\" deux nombres complexes non nuls, alors :\n

  • \"\arg(zz')\equiv\arg(z)+\arg(z')\,[2\pi]\" <\/li>\n
  • \"\arg(\dfrac{z}{z'})\equiv\arg(z)-\arg(z')\,[2\pi]\" <\/li>\n
  • \"\arg(\overline{z})=\arg\left(\dfrac{1}{z}\right)\equiv-\arg(z)\,[2\pi]\" <\/li>\n
  • \"\forall n\in\mathbb{Z}\", \"\arg(z^n)\equiv n\arg(z)\,[2\pi]\" <\/li>\n\n\n\n

    D\u00e9monstration<\/h3>\n\n\n\n

    Pas de difficult\u00e9 si on utilise la forme trigonom\u00e9trique. Le dernier point se montre par r\u00e9currence.<\/p>\n\n\n\n

    Exemple<\/h4>\n\n\n\n

    <\/p>\nSoit \"z=\dfrac{1-\i}{\i}\".
    \nOn peut v\u00e9rifier que \"1-\i= e^{-\i\pi/4}\" et \"\i= e^{\i\pi/2}\" (voir point m\u00e9thodologique).\nAlors :

      <\/span>   <\/span>\"\[\displaystyle\arg\left(\dfrac{1-\i}{\i}\right)=\arg(1-\i)-\arg(\i)\,[2\pi]=-\dfrac{\pi}{4}+\dfrac{\pi}{2}\,[2\pi]=\dfrac{\pi}{4}\,[2\pi].\]\"<\/p>\nDe plus \"\displaystyle\left|\dfrac{1-\i}{\i}\right|=\sqrt{2}.\"\nOn en d\u00e9duit donc que :\n\"z=\sqrt{2}e^{\i\pi/4}.\"\n\n\n\n

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    \"livre<\/figure>\n<\/div>\n\n\n\n
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    Cet article est extrait de l’ouvrage Maths MPSI-MP2I. Tout-en-un : cours, m\u00e9thodes, entra\u00eenement et corrig\u00e9s <\/em>(\u00e9ditions Vuibert, juin 2021) <\/em>\u00e9crit par E. Thomas, S. Bellec, G. Boutard. ISBN n\u00b09782311408720<\/em><\/p>\n<\/div>\n<\/div>\n\n\n\n

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    \n 5\/5 - (1 vote) <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

    Vous \u00e9tudiez actuellement les notions li\u00e9es \u00e0 l\u2019argument des nombres complexes ? Gr\u00e2ce \u00e0 ce cours d\u00e9di\u00e9, ma\u00eetrisez (…)<\/p>\n","protected":false},"author":158,"featured_media":164161,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":true,"footnotes":""},"category":[803,810],"tag":[78,345],"class_list":["post-234650","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-apprendre-matiere","category-maths","tag-prepa","tag-prepa-scientifique"],"acf":[],"_links":{"self":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/234650","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/users\/158"}],"replies":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/comments?post=234650"}],"version-history":[{"count":0,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/posts\/234650\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media\/164161"}],"wp:attachment":[{"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/media?parent=234650"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/category?post=234650"},{"taxonomy":"tag","embeddable":true,"href":"https:\/\/sherpas.com\/blog\/wp-json\/wp\/v2\/tag?post=234650"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}