// Numbas version: finer_feedback_settings
{"name": "Shaheen's copy of De Moivre's Theorem: Positive Powers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Shaheen's copy of De Moivre's Theorem: Positive Powers", "tags": [], "metadata": {"description": "<p>Find modulus and argument of two complex numbers. </p><p>Then use De Moivre's Theorem to find powers of the complex numbers.</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>Use de Moivre's theorem to write the following complex number in the form $a+bi$.</p>\n<p>Remember that the argument of a complex numbers lies in the range $-\\pi \\lt \\theta \\le \\pi$.</p>\n<p><span style=\"text-decoration: underline;\">Important</span></p>\n<p>When calculating the final answer <strong>in part (iii),&nbsp;</strong><strong>you must use whole number values (integers)</strong>, otherwise the final answer will not be marked as correct.</p>", "advice": "<p>Given a complex number $z=r(\\cos(\\theta)+i\\sin(\\theta))$ de Moivre's theorem states that $z^n=r^n(\\cos(n\\theta)+i\\sin(n\\theta))$ for an integer power $n$.<br/>So if we know the modulus $r$ and the argument $\\theta$ for $z$ then the theorem provides a way of calculating $z^n$.</p>\n<p>As usual, you must be careful that the argument is calculated correctly by paying attention to the quadrant of the complex plane in which the complex number lies.</p>\n<p>Also remember that for this question, arguments of complex numbers lie in the range $-\\pi \\lt \\theta \\le \\pi$.</p>\n<p>With the above in mind we can now answer the questions:</p>\n<p>a)</p>\n<h4>Modulus</h4>\n<p>\\[ \\begin{eqnarray*} |\\var{z1}|&amp;=&amp;\\sqrt{(\\var{a1})^2+(\\var{b1})^2}\\\\ &amp;=&amp; \\var{abs(z1)}\\\\ &amp;=&amp;\\var{ans1} \\end{eqnarray*} \\] <strong>to 3 decimal places</strong>.</p>\n<p>Note that $r^{\\var{n2}}=|(\\var{z1})^{\\var{n2}}| =\\var{abs(z1)}^{\\var{n2}}=\\var{abs(z1)^n2}$ which we will use in the calculation for $(\\var{z1})^{\\var{n2}}$</p>\n<h4>Argument</h4>\n<p>{m1}.<br/>Hence we see that: <br/>\\[\\begin{eqnarray*} \\arg(\\var{z1}) &amp;=&amp; \\var{arg(z1)}\\\\ &amp;=&amp; \\var{arg1}\\; \\mbox{radians} \\end{eqnarray*} \\] <strong>to 3 decimal places</strong>.</p>\n<p>We have $\\arg((\\var{z1})^{\\var{n2}})=\\var{n2}\\times \\var{arg(z1)} = \\var{n2*arg(z1)}$ <strong>radians</strong>.</p>\n<p>Hence we have \\[\\begin{eqnarray*}(\\var{z1})^{\\var{n2}} &amp;=&amp; \\var{abs(z1)^n2}(\\cos(\\var{n2*arg(z1)})+\\sin(\\var{n2*arg(z1)})i)\\\\ &amp;=&amp; \\var{abs(z1)^n2}\\cos(\\var{n2*arg(z1)})+\\var{abs(z1)^n2}\\times\\sin(\\var{n2*arg(z1)})i\\\\ &amp;=&amp; \\simplify[std]{{a3}+{b3}i}. \\end{eqnarray*} \\]&nbsp;</p>\n<p>b)</p>\n<h4>Modulus</h4>\n<p>\\[ \\begin{eqnarray*} |\\var{z2}|&amp;=&amp;\\sqrt{(\\var{a2})^2+(\\var{b2})^2}\\\\ &amp;=&amp; \\var{abs(z2)}\\\\ &amp;=&amp;\\var{ans2} \\end{eqnarray*} \\] <strong>to 3 decimal places</strong>.</p>\n<p>Note that $r^{\\var{n4}}=|(\\var{z2})^{\\var{n4}}| =\\var{abs(z2)}^{\\var{n4}}=\\var{abs(z2)^n4}$ which we will use in the calculation for $(\\var{z2})^{\\var{n4}}$</p>\n<h4>Argument</h4>\n<p>{m2}.<br/>Hence we see that: <br/>\\[\\begin{eqnarray*} \\arg(\\var{z2}) &amp;=&amp; \\var{arg(z2)}\\\\ &amp;=&amp; \\var{arg2}\\; \\mbox{radians} \\end{eqnarray*} \\] <strong>to 3 decimal places</strong>.</p>\n<p>We have $\\arg((\\var{z2})^{\\var{n4}})=\\var{n4}\\times \\var{arg(z2)} = \\var{n4*arg(z2)}$ ra<strong>dians</strong>.</p>\n<p>Hence we have \\[\\begin{eqnarray*}(\\var{z2})^{\\var{n4}} &amp;=&amp; \\var{abs(z2)^n4}(\\cos(\\var{n4*arg(z2)})+\\sin(\\var{n4*arg(z2)})i)\\\\ &amp;=&amp; \\var{abs(z2)^n4}\\cos(\\var{n4*arg(z2)})+\\var{abs(z2)^n4}\\times\\sin(\\var{n4*arg(z2)})i\\\\ &amp;=&amp; \\simplify[std]{{a4}+{b4}i}. \\end{eqnarray*} \\]&nbsp;</p>\n<p>&nbsp;</p>", "rulesets": 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