// Numbas version: exam_results_page_options
{"name": "Divis\u00e3o de n\u00fameros complexos I", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Divis\u00e3o de n\u00fameros complexos I", "tags": ["checked2015", "complex numbers", "conjugate of a complex number", "division of complex numbers", "inverse of complex numbers", "multiplication of complex numbers"], "metadata": {"description": "<p>Inverse and division of complex numbers. &nbsp;Four parts.</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>Apresente na forma $a+bi$.</p>\n<p>&nbsp;Escreva $a$ e $b$ como fra&ccedil;&otilde;es ou inteiros, n&atilde;o como decimais.</p>", "advice": "<p>A divis&atilde;o de dois complexos pode ser realizada atrav&eacute;s da multiplica&ccedil;&atilde;o do numerador e do denominador pelo conjugado do denominador.</p>\n<p><br/>Suponha que \\[ z = \\frac{a+bi}{c+di},\\;\\; c+di \\neq 0\\] ent&atilde;o temos:<br/>\\[\\begin{eqnarray*} z&amp;=&amp;\\frac{a+bi}{c+di}\\\\ &amp;=&amp;\\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\\\ &amp;=&amp;\\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\\\\ &amp;=&amp;\\frac{ac+bd}{c^2+d^2}+\\frac{bc-ad}{c^2+d^2}i \\end{eqnarray*} \\]<br/>Apesar desta f&oacute;rmula, a melhor maneira de calcular a divis&atilde;o &eacute; lembrar que &eacute; necess&aacute;rio multiplicar o numerador e o denominador pelo conjugado.<br/>(a)<br/>\\[\\begin{eqnarray*}\\simplify[std]{{c1}/{z1}} &amp;=&amp;\\simplify[std]{({c1}*{conj(z1)})/({z1}*{conj(z1)})}\\\\ &amp;=&amp;\\simplify[std]{{c1*conj(z1)}/{abs(z1)^2}}\\\\ &amp;=&amp; \\simplify[std]{{c1*re(z1)}/{abs(z1)^2}-{c1*im(z1)}/{abs(z1)^2}*i} \\end{eqnarray*} \\]<br/>(b)<br/>\\[\\begin{eqnarray*}\\simplify[std]{{c2}/{z2}} &amp;=&amp;\\simplify[std]{({c2}*{conj(z2)})/({z2}*{conj(z2)})}\\\\ &amp;=&amp;\\simplify[std]{{c2*conj(z2)}/{abs(z2)^2}}\\\\ &amp;=&amp; \\simplify[std]{{c2*re(z2)}/{abs(z2)^2}-{c2*im(z2)}/{abs(z2)^2}*i} \\end{eqnarray*} \\]<br/>(c)<br/>\\[\\begin{eqnarray*}\\simplify[std]{{z1}/{z3}} &amp;=&amp;\\simplify[std]{({z1}*{conj(z3)})/({z3}*{conj(z3)})}\\\\ &amp;=&amp;\\simplify[std]{{z1*conj(z3)}/{abs(z3)^2}}\\\\ &amp;=&amp; \\simplify[std]{{re(z1*conj(z3))}/{abs(z3)^2}+{im(z1*conj(z3))}/{abs(z3)^2}*i} \\end{eqnarray*} \\]<br/>(d)<br/>\\[\\begin{eqnarray*}\\simplify[std]{{z3}/{z2}} &amp;=&amp;\\simplify[std]{({z3}*{conj(z2)})/({z2}*{conj(z2)})}\\\\ &amp;=&amp;\\simplify[std]{{z3*conj(z2)}/{abs(z2)^2}}\\\\ &amp;=&amp; \\simplify[std]{{re(z3*conj(z2))}/{abs(z2)^2}+{im(z3*conj(z2))}/{abs(z2)^2}*i} \\end{eqnarray*} \\]</p>", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus", "!collectLikeFractions"]}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"s1": {"name": 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