{"id":9278,"date":"2012-05-22T01:23:17","date_gmt":"2012-05-22T00:23:17","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=9278"},"modified":"2021-12-29T15:29:03","modified_gmt":"2021-12-29T15:29:03","slug":"mostre-que-6","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=9278","title":{"rendered":"Mostre que"},"content":{"rendered":"<p><ul id='GTTabs_ul_9278' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_9278' class='GTTabs_curr'><a  id=\"9278_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_9278' ><a  id=\"9278_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_9278'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Mostre que $${\\left( {1 + \\sqrt 3 i} \\right)^n} + {\\left( {1 &#8211; \\sqrt 3 i} \\right)^n} = {2^{n + 1}}\\operatorname{c} os\\left( {\\frac{{n\\pi }}{3}} \\right)$$ para todo o $n \\in \\mathbb{N}$.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_9278' onClick='GTTabs_show(1,9278)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_9278'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>Ora,<\/p>\n<p>$$\\begin{array}{*{20}{l}}<br \/>\n{{{\\left( {1 + \\sqrt 3 i} \\right)}^n} + {{\\left( {1 &#8211; \\sqrt 3 i} \\right)}^n}}&amp; = &amp;{{{\\left[ {2\\left( {\\frac{1}{2} + \\frac{{\\sqrt 3 }}{2}i} \\right)} \\right]}^n} + {{\\left[ {2\\left( {\\frac{1}{2} &#8211; \\frac{{\\sqrt 3 }}{2}i} \\right)} \\right]}^n}} \\\\<br \/>\n{}&amp; = &amp;{{{\\left( {2\\operatorname{cis} \\frac{\\pi }{3}} \\right)}^n} + {{\\left( {2\\operatorname{cis} \\left( { &#8211; \\frac{\\pi }{3}} \\right)} \\right)}^n}} \\\\<br \/>\n{}&amp; = &amp;{{2^n}\\operatorname{cis} \\left( {\\frac{{n\\pi }}{3}} \\right) + {2^n}\\operatorname{cis} \\left( { &#8211; \\frac{{n\\pi }}{3}} \\right)} \\\\<br \/>\n{}&amp; = &amp;{{2^n}\\left[ {\\cos \\left( {\\frac{{n\\pi }}{3}} \\right) + i\\operatorname{sen} \\left( {\\frac{{n\\pi }}{3}} \\right) + \\cos \\left( { &#8211; \\frac{{n\\pi }}{3}} \\right) + i\\operatorname{sen} \\left( { &#8211; \\frac{{n\\pi }}{3}} \\right)} \\right]} \\\\<br \/>\n{}&amp; = &amp;{{2^n}\\left[ {\\cos \\left( {\\frac{{n\\pi }}{3}} \\right) + i\\operatorname{sen} \\left( {\\frac{{n\\pi }}{3}} \\right) + \\cos \\left( {\\frac{{n\\pi }}{3}} \\right) &#8211; i\\operatorname{sen} \\left( {\\frac{{n\\pi }}{3}} \\right)} \\right]} \\\\<br \/>\n{}&amp; = &amp;{{2^n} \\times 2\\cos \\left( {\\frac{{n\\pi }}{3}} \\right)} \\\\<br \/>\n{}&amp; = &amp;{{2^{n + 1}}\\cos \\left( {\\frac{{n\\pi }}{3}} \\right)}<br \/>\n\\end{array}$$<\/p>\n<p>Portanto, $${\\left( {1 + \\sqrt 3 i} \\right)^n} + {\\left( {1 &#8211; \\sqrt 3 i} \\right)^n} = {2^{n + 1}}\\operatorname{c} os\\left( {\\frac{{n\\pi }}{3}} \\right),\\forall n \\in \\mathbb{N}$$<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_9278' onClick='GTTabs_show(0,9278)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Mostre que $${\\left( {1 + \\sqrt 3 i} \\right)^n} + {\\left( {1 &#8211; \\sqrt 3 i} \\right)^n} = {2^{n + 1}}\\operatorname{c} os\\left( {\\frac{{n\\pi }}{3}} \\right)$$ para todo o $n \\in \\mathbb{N}$.&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19170,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,310],"tags":[427,314,18],"series":[],"class_list":["post-9278","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-forma-trigonometrica","tag-numeros-complexos"],"views":2668,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat61.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9278","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=9278"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9278\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=9278"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=9278"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=9278"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=9278"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}