{"id":9073,"date":"2012-05-15T16:21:17","date_gmt":"2012-05-15T15:21:17","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=9073"},"modified":"2022-01-27T00:24:05","modified_gmt":"2022-01-27T00:24:05","slug":"cardan-e-a-nocao-de-numero-complexo","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=9073","title":{"rendered":"Cardan e a no\u00e7\u00e3o de n\u00famero complexo"},"content":{"rendered":"<p><ul id='GTTabs_ul_9073' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_9073' class='GTTabs_curr'><a  id=\"9073_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_9073' ><a  id=\"9073_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_9073'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<div id=\"attachment_9080\" style=\"width: 240px\" class=\"wp-caption alignright\"><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-9080\" data-attachment-id=\"9080\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=9080\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan.jpg\" data-orig-size=\"280,364\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Gerolamo Cardano\" data-image-description=\"\" data-image-caption=\"&lt;p&gt;Gerolamo Cardano (1501 &amp;#8211; 1576)&lt;\/p&gt;\n\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan.jpg\" class=\"  wp-image-9080 size-medium\" title=\"Gerolamo Cardano\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan-230x300.jpg\" alt=\"\" width=\"230\" height=\"300\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan-230x300.jpg 230w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan-115x150.jpg 115w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/cardan.jpg 280w\" sizes=\"auto, (max-width: 230px) 100vw, 230px\" \/><\/a><p id=\"caption-attachment-9080\" class=\"wp-caption-text\">Gerolamo Cardano (1501 &#8211; 1576)<\/p><\/div>\n<p>No s\u00e9c. XVI, ao procurar decompor 10 em dois n\u00fameros cujo produto fosse 40, o Matem\u00e1tico <a title=\"Cardan\" href=\"http:\/\/en.wikipedia.org\/wiki\/Gerolamo_Cardano\" target=\"_blank\" rel=\"noopener\">Cardan<\/a> fez uma primeira abordagem \u00e0 no\u00e7\u00e3o de n\u00famero complexo, tendo, no entanto, qualificado de &#8220;sofisticadas&#8221; as ra\u00edzes quadradas de n\u00fameros negativos e de &#8220;subtil e in\u00fatil&#8221; o resultado a que chegou.<\/p>\n<ol>\n<li>a) Verifique que \u00e9 imposs\u00edvel encontrar dois n\u00fameros reais cuja soma seja 10 e cujo produto seja 40.\n<p>b) Encontre os n\u00fameros complexos ${z_1}$ e ${z_2}$ que satisfazem estas condi\u00e7\u00f5es.<br \/>\nDesigne por ${z_1}$ o que tem coeficiente da parte imagin\u00e1ria negativo.<\/p>\n<p>c) Marque no plano complexo os afixos ${M_1}$ e ${M_2}$ desses complexos.<\/p>\n<\/li>\n<li>a) Resolva, em $\\mathbb{C}$, a equa\u00e7\u00e3o $$16{z^4} &#8211; {z^2} &#8211; 15 = 0$$\n<p>b) Marque no plano de Argand as imagens das solu\u00e7\u00f5es.<\/p>\n<p>c) Seja A o afixo de $\\frac{{ &#8211; \\sqrt {15} }}{4}i$ e B o afixo de 1.<br \/>\nMostre que A, B e ${M_2}$ s\u00e3o colineares.<\/p>\n<p>d) Encontre o afixo M do complexo $z$ tal que $\\left[ {A{M_2}M{M_1}} \\right]$ seja um paralelogramo. Construa-o.<\/p>\n<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_9073' onClick='GTTabs_show(1,9073)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_9073'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>a) Como sabemos, se ${x_1}$ e ${x_2}$ s\u00e3o duas solu\u00e7\u00f5es reais de uma equa\u00e7\u00e3o do 2.\u00ba grau, ent\u00e3o ela pode ser escrita na forma $${x^2} &#8211; Sx + P = 0$$ com $S = {x_1} + {x_2}$ e $P = {x_1} \\times {x_2}$. [<a href=\"https:\/\/www.acasinhadamatematica.pt\/?p=8571\" target=\"_blank\" rel=\"noopener\">ver mais<\/a> (prova ao cuidado do leitor)]\n<p>Assim, em $\\mathbb{R}$, temos: $$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; 10x + 40 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{10 \\pm \\sqrt {100 &#8211; 160} }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{10 \\pm \\sqrt { &#8211; 60} }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x \\in \\emptyset }<br \/>\n\\end{array}$$<br \/>\nPortanto, n\u00e3o existem dois n\u00fameros reais cuja soma seja 10 e cujo produto seja 40.<\/p>\n<p>b) Em $\\mathbb{C}$, temos: $$\\begin{array}{*{20}{l}}<br \/>\n{{z^2} &#8211; 10z + 40 = 0}&amp; \\Leftrightarrow &amp;{z = \\frac{{10 \\pm \\sqrt {100 &#8211; 160} }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{z = \\frac{{10 \\pm \\sqrt { &#8211; 60} }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{z = 5 &#8211; \\sqrt {15} i}&amp; \\vee &amp;{z = 5 + \\sqrt {15} i}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<br \/>\nPortanto, ${z_1} = 5 &#8211; \\sqrt {15} i$ e ${z_2} = 5 + \\sqrt {15} i$.<\/p>\n<p>c) Ver abaixo.<\/p>\n<\/li>\n<li>a) Resolvendo a equa\u00e7\u00e3o, temos: $$\\begin{array}{*{20}{l}}<br \/>\n{16{z^4} &#8211; {z^2} &#8211; 15 = 0}&amp; \\Leftrightarrow &amp;{{z^2} = \\frac{{1 \\pm \\sqrt {1 + 960} }}{{32}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{z^2} = \\frac{{1 \\pm 31}}{{32}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{{z^2} = 1}&amp; \\vee &amp;{{z^2} =\u00a0 &#8211; \\frac{{15}}{{16}}}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{z =\u00a0 &#8211; 1}&amp; \\vee &amp;{z = 1}&amp; \\vee &amp;{z =\u00a0 &#8211; \\frac{{\\sqrt {15} }}{4}i}&amp; \\vee &amp;{z = \\frac{{\\sqrt {15} }}{4}i}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/p>\n<p>b) \u00a0Ver abaixo.<\/p>\n<p>c) Sejam: $$\\begin{array}{*{20}{l}}<br \/>\n{{M_1}\\left( {5, &#8211; \\sqrt {15} } \\right)}&amp;;&amp;{{M_2}\\left( {5,\\sqrt {15} } \\right)}&amp;;&amp;{A\\left( {0, &#8211; \\frac{{\\sqrt {15} }}{4}} \\right)}&amp;;&amp;{B\\left( {1,0} \\right)}&amp;;&amp;{C\\left( { &#8211; 1,0} \\right)}&amp;{\\text{e}}&amp;{D\\left( {0,\\frac{{\\sqrt {15} }}{4}} \\right)}<br \/>\n\\end{array}$$<br \/>\nComo $\\overrightarrow {AB}\u00a0 = \\left( {1,\\frac{{\\sqrt {15} }}{4}} \\right)$ e $\\overrightarrow {B{M_2}}\u00a0 = \\left( {4,\\sqrt {15} } \\right)$, ent\u00e3o $\\overrightarrow {B{M_2}}\u00a0 = 4\\overrightarrow {AB} $. Logo, sendo os vetores colineares, os tr\u00eas pontos considerados s\u00e3o colineares.<\/p>\n<p>d) Para que que $\\left[ {A{M_2}M{M_1}} \\right]$ seja um paralelogramo, ent\u00e3o $$\\begin{array}{*{20}{l}}<br \/>\nM&amp; = &amp;{{M_2} + \\overrightarrow {A{M_1}} } \\\\<br \/>\n{}&amp; = &amp;{\\left( {5,\\sqrt {15} } \\right) + \\left( {5, &#8211; \\frac{{3\\sqrt {15} }}{4}} \\right)} \\\\<br \/>\n{}&amp; = &amp;{\\left( {10,\\frac{{\\sqrt {15} }}{4}} \\right)}<br \/>\n\\end{array}$$<\/p>\n<\/li>\n<\/ol>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"9091\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=9091\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44.png\" data-orig-size=\"595,373\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Plano de Argand\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44.png\" class=\"aligncenter size-full wp-image-9091\" title=\"Plano de Argand\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44.png\" alt=\"\" width=\"595\" height=\"373\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44.png 595w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44-300x188.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44-150x94.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12pag139-44-400x250.png 400w\" sizes=\"auto, (max-width: 595px) 100vw, 595px\" \/><\/a><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_9073' onClick='GTTabs_show(0,9073)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado No s\u00e9c. XVI, ao procurar decompor 10 em dois n\u00fameros cujo produto fosse 40, o Matem\u00e1tico Cardan fez uma primeira abordagem \u00e0 no\u00e7\u00e3o de n\u00famero complexo, tendo, no entanto, qualificado de&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":21079,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,310],"tags":[427,18,313],"series":[],"class_list":["post-9073","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-12--ano","category-aplicando","category-numeros-complexos-12--ano","tag-12-o-ano","tag-numeros-complexos","tag-plano-de-argand"],"views":1871,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/05\/12V3Pag140-44_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9073","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=9073"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9073\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/21079"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=9073"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=9073"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=9073"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=9073"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}