Resolu\u00e7\u00e3o<\/b><\/span><\/p>\nSeja $z = {\\cos ^2}\\theta\u00a0 + \\frac{i}{2}\\operatorname{sen} \\left( {2\\theta } \\right)$ e $\\theta\u00a0 \\in \\left] { – \\pi ,\\pi } \\right[$.<\/p>\n
Exerc\u00edcio 1<\/strong><\/span><\/p>\nO m\u00f3dulo de $z$ \u00e9:<\/p>\n
[A]<\/strong> $\\cos \\theta $, qualquer que seja $\\theta $;<\/p>\n[B]<\/strong> $\\cos \\theta $, se $\\theta\u00a0 \\in \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]$;<\/p>\n[C]<\/strong> $\\cos \\left( {\\theta\u00a0 + \\pi } \\right)$, se $\\theta\u00a0 \\notin \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]$.<\/p>\n\nOra, $$\\begin{array}{*{20}{l}}
\n{\\left| {\\text{z}} \\right|}& = &{\\sqrt {{{\\left( {{{\\cos }^2}\\theta } \\right)}^2} + {{\\left( {\\frac{1}{2}\\operatorname{sen} \\left( {2\\theta } \\right)} \\right)}^2}} } \\\\
\n{}& = &{\\sqrt {{{\\cos }^4}\\theta\u00a0 + \\frac{1}{4}{{\\operatorname{sen} }^2}\\left( {2\\theta } \\right)} } \\\\
\n{}& = &{\\sqrt {{{\\cos }^4}\\theta\u00a0 + \\frac{1}{4}{{\\left( {\\operatorname{sen} \\left( {2\\theta } \\right)} \\right)}^2}} } \\\\
\n{}& = &{\\sqrt {{{\\cos }^4}\\theta\u00a0 + \\frac{1}{4}{{\\left( {2\\operatorname{sen} \\theta \\cos \\theta } \\right)}^2}} } \\\\
\n{}& = &{\\sqrt {{{\\cos }^4}\\theta\u00a0 + \\frac{1}{4} \\times 4{{\\operatorname{sen} }^2}\\theta\u00a0 \\times {{\\cos }^2}\\theta } } \\\\
\n{}& = &{\\sqrt {{{\\cos }^2}\\theta \\left( {{{\\cos }^2}\\theta\u00a0 + {{\\operatorname{sen} }^2}\\theta } \\right)} } \\\\
\n{}& = &{\\sqrt {{{\\cos }^2}\\theta } } \\\\
\n{}& = &{\\left| {\\cos \\theta } \\right|} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{\\cos \\theta }& \\Leftarrow &{\\theta\u00a0 \\in \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]} \\\\
\n{\\cos \\left( {\\pi\u00a0 + \\theta } \\right)}& \\Leftarrow &{\\theta\u00a0 \\in \\left( {\\left] { – \\pi ,\\pi } \\right[\\backslash \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]} \\right)}
\n\\end{array}} \\right.}
\n\\end{array}$$<\/p>\n
Logo, s\u00e3o corretas as respostas B<\/strong> e C<\/strong>.<\/p>\n<\/blockquote>\nExerc\u00edcio 2<\/strong><\/span><\/p>\nUm argumento de $z$ \u00e9:<\/p>\n
[A]<\/strong> $\\theta $, qualquer que seja $\\theta $;<\/p>\n[B]<\/strong> $2\\theta $:<\/p>\n[C]<\/strong> $\\theta\u00a0 + \\pi $, se $\\theta\u00a0 \\notin \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]$.<\/p>\n\nOra, $$\\begin{array}{*{20}{l}}
\nz& = &{{{\\cos }^2}\\theta\u00a0 + \\frac{i}{2}\\operatorname{sen} \\left( {2\\theta } \\right)} \\\\
\n{}& = &{{{\\cos }^2}\\theta\u00a0 + \\frac{i}{2} \\times 2\\operatorname{sen} \\theta \\cos \\theta } \\\\
\n{}& = &{\\cos \\theta \\left( {\\cos \\theta\u00a0 + i\\operatorname{sen} \\theta } \\right)}
\n\\end{array}$$<\/p>\n
Como ${\\cos ^2}\\theta\u00a0 \\geqslant 0,\\forall \\left] { – \\pi ,\\pi } \\right[$, ent\u00e3o $\\arg z \\in \\left[ { – \\frac{\\pi }{2} + 2k\\pi ,\\frac{\\pi }{2} + 2k\\pi } \\right],k \\in \\mathbb{Z}$<\/p>\n
Assim:<\/p>\n
\n- um argumento de $z$ \u00e9 $\\theta $ se ${\\theta\u00a0 \\in \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]}$.<\/li>\n
- um argumento de $z$ \u00e9 $\\theta\u00a0 + \\pi $ se ${\\theta\u00a0 \\in \\left( {\\left] { – \\pi ,\\pi } \\right[\\backslash \\left[ { – \\frac{\\pi }{2},\\frac{\\pi }{2}} \\right]} \\right)}$<\/li>\n<\/ul>\n
Logo, apenas a resposta C<\/strong> \u00e9 correta.<\/p>\n<\/blockquote>\nExerc\u00edcio 3<\/strong><\/span><\/p>\n${z^2}$ \u00e9 igual a:<\/p>\n
[A]<\/strong> ${\\cos ^4}\\theta\u00a0 – \\frac{1}{4}{\\operatorname{sen} ^2}2\\theta\u00a0 + i\\operatorname{sen} 2\\theta {\\cos ^2}\\theta $;<\/p>\n[B]<\/strong> ${\\cos ^2}\\theta \\operatorname{cis} \\left( {2\\theta } \\right)$;<\/p>\n[C]<\/strong> ${\\cos ^2}\\left( {\\theta\u00a0 + \\pi } \\right)\\operatorname{cis} \\left[ {2\\left( {\\theta\u00a0 + \\pi } \\right)} \\right]$.<\/p>\n\n$$\\begin{array}{*{20}{l}}
\n{{z^2}}& = &{{{\\left( {{{\\cos }^2}\\theta\u00a0 + \\frac{i}{2}\\operatorname{sen} \\left( {2\\theta } \\right)} \\right)}^2}}&{}&{} \\\\
\n{}& = &{{{\\cos }^4}\\theta\u00a0 – \\frac{1}{4}{{\\operatorname{sen} }^2}\\left( {2\\theta } \\right) + i\\operatorname{sen} \\left( {2\\theta } \\right){{\\cos }^2}\\theta }&{}&{{\\text{[A]}}} \\\\
\n{}& = &{{{\\cos }^4}\\theta\u00a0 – \\frac{1}{4}{{\\left( {2\\operatorname{sen} \\theta \\cos \\theta } \\right)}^{\\text{2}}} + i\\operatorname{sen} \\left( {2\\theta } \\right){{\\cos }^2}\\theta }&{}&{} \\\\
\n{}& = &{{{\\cos }^2}\\theta \\left( {{{\\cos }^2}\\theta\u00a0 – {{\\operatorname{sen} }^2}\\theta } \\right) + i\\operatorname{sen} \\left( {2\\theta } \\right){{\\cos }^2}\\theta }&{}&{} \\\\
\n{}& = &{{{\\cos }^2}\\theta \\cos \\left( {2\\theta } \\right) + i\\operatorname{sen} \\left( {2\\theta } \\right){{\\cos }^2}\\theta }&{}&{} \\\\
\n{}& = &{{{\\cos }^2}\\theta \\left( {\\cos \\left( {2\\theta } \\right) + i\\operatorname{sen} \\left( {2\\theta } \\right)} \\right)}&{}&{} \\\\
\n{}& = &{{{\\cos }^2}\\theta \\operatorname{cis} \\left( {2\\theta } \\right)}&{}&{{\\text{[B]}}} \\\\
\n{}& = &{{{\\cos }^2}\\left( {\\theta\u00a0 + \\pi } \\right)\\operatorname{cis} \\left( {2\\theta\u00a0 + 2\\pi } \\right)}&{}&{} \\\\
\n{}& = &{{{\\cos }^2}\\left( {\\theta\u00a0 + \\pi } \\right)\\operatorname{cis} \\left( {2\\left( {\\theta\u00a0 + \\pi } \\right)} \\right)}&{}&{{\\text{[C]}}}
\n\\end{array}$$<\/p>\n
Logo, todas as respostas s\u00e3o corretas.<\/p>\n<\/blockquote>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"Enunciado Resolu\u00e7\u00e3o Enunciado Em cada uma das al\u00edneas seguintes, uma ou v\u00e1rias respostas est\u00e3o corretas. Indique quais. Seja $z = {\\cos ^2}\\theta\u00a0 + \\frac{i}{2}\\operatorname{sen} \\left( {2\\theta } \\right)$ e $\\theta\u00a0 \\in \\left] { –...<\/p>\n","protected":false},"author":1,"featured_media":19570,"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],"series":[],"class_list":["post-9239","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"],"views":1425,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat189.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9239","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=9239"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/9239\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19570"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=9239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=9239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=9239"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=9239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}