{"id":9134,"date":"2012-05-20T19:06:25","date_gmt":"2012-05-20T18:06:25","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=9134"},"modified":"2021-12-29T02:39:41","modified_gmt":"2021-12-29T02:39:41","slug":"represente-na-forma-trigonometrica-os-numeros-complexos","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=9134","title":{"rendered":"Represente na forma trigonom\u00e9trica os n\u00fameros complexos"},"content":{"rendered":"<p><ul id='GTTabs_ul_9134' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_9134' class='GTTabs_curr'><a  id=\"9134_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_9134' ><a  id=\"9134_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_9134'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Represente na forma trigonom\u00e9trica os n\u00fameros complexos:<\/p>\n<ol>\n<li>$z = 3 + 3i$<\/li>\n<li>$z =\u00a0 &#8211; 1 &#8211; i$<\/li>\n<li>$z = 4i$<\/li>\n<li>$z =\u00a0 &#8211; 0,6i$<\/li>\n<li>$z =\u00a0 &#8211; \\frac{{\\sqrt 2 }}{2}$<\/li>\n<li>$z = \\sqrt 2\u00a0 &#8211; \\sqrt 6 i$<\/li>\n<li>$z =\u00a0 &#8211; 3 + \\sqrt 3 i$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_9134' onClick='GTTabs_show(1,9134)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_9134'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>$$z = 3 + 3i$$<br \/>\nA localiza\u00e7\u00e3o do afixo do n\u00famero complexo (no primeiro quadrante, sobre a bissetriz dos quadrantes \u00edmpares, a $3\\sqrt 2 $ unidades da origem do referencial) permite-o escrever imediatamente na forma trigonom\u00e9trica: $z = 3\\sqrt 2 \\operatorname{cis} \\frac{\\pi }{4}$.<\/p>\n<p>De qualquer forma, vem:<br \/>\n$$\\left| z \\right| = \\sqrt {{3^2} + {3^2}}\u00a0 = 3\\sqrt 2 $$<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{3}{{3\\sqrt 2 }}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{3}{{3\\sqrt 2 }}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{{\\sqrt 2 }}{2}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{{\\sqrt 2 }}{2}}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\frac{\\pi }{4}}<br \/>\n\\end{array}.$$<br \/>\nLogo, $z = 3\\sqrt 2 \\operatorname{cis} \\frac{\\pi }{4}$.<\/p>\n<\/li>\n<li>$$z =\u00a0 &#8211; 1 &#8211; i$$<br \/>\nA localiza\u00e7\u00e3o do afixo do n\u00famero complexo (no\u00a0terceiro quadrante, sobre a bissetriz dos quadrantes \u00edmpares, a $\\sqrt 2 $ unidades da origem do referencial) permite-o escrever imediatamente na forma trigonom\u00e9trica: $z = \\sqrt 2 \\operatorname{cis} \\frac{{5\\pi }}{4}$.<\/p>\n<p>De qualquer forma, vem:<br \/>\n$$\\left| z \\right| = \\sqrt {{{\\left( { &#8211; 1} \\right)}^2} + {{\\left( { &#8211; 1} \\right)}^2}}\u00a0 = \\sqrt 2 $$<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{{ &#8211; 1}}{{\\sqrt 2 }}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{{ &#8211; 1}}{{\\sqrt 2 }}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 =\u00a0 &#8211; \\frac{{\\sqrt 2 }}{2}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 =\u00a0 &#8211; \\frac{{\\sqrt 2 }}{2}}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\frac{{5\\pi }}{4}}<br \/>\n\\end{array}.$$<br \/>\nLogo, $z = \\sqrt 2 \\operatorname{cis} \\frac{{5\\pi }}{4}$.<\/p>\n<\/li>\n<li>$$z = 4i$$<br \/>\nA localiza\u00e7\u00e3o do afixo do n\u00famero complexo (sobre o semieixo positivo $Oy$ a $4$ unidades da origem do referencial) permite-o escrever imediatamente na forma trigonom\u00e9trica: $z = 4\\operatorname{cis} \\frac{\\pi }{2}$.<\/p>\n<p>De qualquer forma, vem:<br \/>\n$$\\left| z \\right| = \\sqrt {{0^2} + {4^2}}\u00a0 = 4$$<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{0}{4}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{4}{4}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = 0} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = 1}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\frac{\\pi }{2}}<br \/>\n\\end{array}.$$<br \/>\nLogo, $z = 4\\operatorname{cis} \\frac{\\pi }{2}$.<\/p>\n<\/li>\n<li>$$z =\u00a0 &#8211; 0,6i$$<br \/>\nA localiza\u00e7\u00e3o do afixo do n\u00famero complexo (sobre o semieixo negativo $Oy$ a $0,6$ unidades da origem do referencial) permite-o escrever imediatamente na forma trigonom\u00e9trica: $z = 0,6\\operatorname{cis} \\frac{{3\\pi }}{2}$.<\/p>\n<p>De qualquer forma, vem:<br \/>\n$$\\left| z \\right| = \\sqrt {{0^2} + {{\\left( { &#8211; 0,6} \\right)}^2}}\u00a0 = 0,6$$<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{0}{{0,6}}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{{ &#8211; 0,6}}{{0,6}}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = 0} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 =\u00a0 &#8211; 1}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\frac{{3\\pi }}{2}}<br \/>\n\\end{array}.$$<br \/>\nLogo, $z = 0,6\\operatorname{cis} \\frac{{3\\pi }}{2}$.<\/p>\n<\/li>\n<li>$$z =\u00a0 &#8211; \\frac{{\\sqrt 2 }}{2}$$<br \/>\nA localiza\u00e7\u00e3o do afixo do n\u00famero complexo (sobre o semieixo negativo $Ox$ a\u00a0${\\frac{{\\sqrt 2 }}{2}}$ unidades da origem do referencial) permite-o escrever imediatamente na forma trigonom\u00e9trica: $z = \\frac{{\\sqrt 2 }}{2}\\operatorname{cis} \\pi $.<\/p>\n<p>De qualquer forma, vem:<br \/>\n$$\\left| z \\right| = \\sqrt {{{\\left( { &#8211; \\frac{{\\sqrt 2 }}{2}} \\right)}^2} + {0^2}}\u00a0 = \\frac{{\\sqrt 2 }}{2}$$<br \/>\n$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{{ &#8211; \\frac{{\\sqrt 2 }}{2}}}{{\\frac{{\\sqrt 2 }}{2}}}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{0}{{\\frac{{\\sqrt 2 }}{2}}}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 =\u00a0 &#8211; 1} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = 0}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\pi }<br \/>\n\\end{array}.$<br \/>\nLogo, $z = \\frac{{\\sqrt 2 }}{2}\\operatorname{cis} \\pi $.<\/p>\n<\/li>\n<li>$$z = \\sqrt 2\u00a0 &#8211; \\sqrt 6 i$$<br \/>\nOra,<br \/>\n$$\\left| z \\right| = \\sqrt {{{\\left( {\\sqrt 2 } \\right)}^2} + {{\\left( { &#8211; \\sqrt 6 } \\right)}^2}}\u00a0 = \\sqrt 8\u00a0 = 2\\sqrt 2 $$<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{{\\sqrt 2 }}{{2\\sqrt 2 }}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{{ &#8211; \\sqrt 6 }}{{2\\sqrt 2 }}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{1}{2}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 =\u00a0 &#8211; \\frac{{\\sqrt 3 }}{2}}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\frac{{2\\pi }}{3}}<br \/>\n\\end{array}.$$<br \/>\nLogo, $z = 2\\sqrt 2 \\operatorname{cis} \\frac{{2\\pi }}{3}$.<\/li>\n<li>$$z =\u00a0 &#8211; 3 + \\sqrt 3 i$$<br \/>\nOra,<br \/>\n$$\\left| z \\right| = \\sqrt {{{\\left( { &#8211; 3} \\right)}^2} + {{\\left( {\\sqrt 3 } \\right)}^2}}\u00a0 = \\sqrt {12}\u00a0 = 2\\sqrt 3 $$<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 = \\frac{{ &#8211; 3}}{{2\\sqrt 3 }}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{{\\sqrt 3 }}{{2\\sqrt 3 }}}<br \/>\n\\end{array}} \\right.}&amp; \\Leftrightarrow &amp;{\\left\\{ {\\begin{array}{*{20}{l}}<br \/>\n{\\cos \\theta\u00a0 =\u00a0 &#8211; \\frac{{\\sqrt 3 }}{2}} \\\\<br \/>\n{\\operatorname{sen} \\theta\u00a0 = \\frac{1}{2}}<br \/>\n\\end{array}} \\right.}&amp;{}&amp;{{\\text{Portanto}}{\\text{, uma solu\u00e7\u00e3o \u00e9\u00a0 }}\\theta\u00a0 = \\frac{{5\\pi }}{6}}<br \/>\n\\end{array}.$$<br \/>\nLogo, $z = 2\\sqrt 3 \\operatorname{cis} \\frac{{5\\pi }}{6}$.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_9134' onClick='GTTabs_show(0,9134)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Represente na forma trigonom\u00e9trica os n\u00fameros complexos: $z = 3 + 3i$ $z =\u00a0 &#8211; 1 &#8211; i$ $z = 4i$ $z =\u00a0 &#8211; 0,6i$ $z =\u00a0 &#8211; \\frac{{\\sqrt 2 }}{2}$&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19571,"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-9134","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":3173,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat190.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9134","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=9134"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9134\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19571"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=9134"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=9134"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=9134"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=9134"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}