// Numbas version: finer_feedback_settings
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Four parts.</p>\n<p>Original: https://numbas.mathcentre.ac.uk/question/11788/arithmetics-of-complex-numbers-v/ by Newcastle University Mathematics and Statistics</p>\n<p>Tramslated to German</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>Schreiben Sie die folgenden komplexen Zahlen $z$ in der Form $a+bi$ mit reellen Zahlen $a$ und $b$.</p>\n<p>Geben Sie $a$ und $b$ als Bruchzahlen oder als ganze Zahlen an.</p>\n<p>Je nachdem, wieviel \"Gl&uuml;ck\" Sie mit dem Zufallszahlengenerator haben, sind die Zahlen bei dieser Aufgabe ziemlich gro&szlig;. Lassen Sie gegebenenfalls eine neue Aufgabe erzeugen oder nehmen Sie f&uuml;r die Rechnungen in den ganzen Zahlen, die auszuf&uuml;hren sind, einen Taschenrechner zur Hand...</p>", "advice": "<p>a)</p>\n<p>F&uuml;r jede komplexe Zahl $z=a+bi$ gilt</p>\n<p>$(a+bi)(a-bi) =a^2+b^2$.</p>\n<p>Also haben wir<br/>\\[\\begin{eqnarray*}z&amp;=&amp;(\\var{z1})^{\\var{c1}}(\\var{conj(z1)})^{\\var{c1}}\\\\ &amp;=&amp;((\\var{z1})(\\var{conj(z1)}))^{\\var{c1}}\\\\ &amp;=&amp;\\simplify[]{({re(z1)}^2+{im(z1)}^2)^{c1}}\\\\ &amp;=&amp;\\var{(re(z1)^2+im(z1)^2)^c1} \\end{eqnarray*}\\]</p>\n<p>b)</p>\n<p>Beachten Sie, dass $(\\var{z2})^4=((\\var{z2})^2)^2$.</p>\n<p>Weil $(\\var{z2})^2=\\simplify[std]{{z2^2}}$, gilt<br/>\\[(\\var{z2})^4=(\\simplify[std]{{z2^2}})^2=\\simplify[std]{{z2^4}}\\]</p>\n<p>c) <br/>Es gilt<br/>\\[ \\begin{eqnarray*} z&amp;=&amp;(\\var{z3})^{\\var{-d1}}\\\\ &amp;=&amp;\\frac{1}{(\\var{z3})^{\\var{d1}}}\\\\ &amp;=&amp;\\frac{(\\var{conj(z3)})^{\\var{d1}}}{(\\var{z3})^{\\var{d1}}(\\var{conj(z3)})^{\\var{d1}}}\\\\ &amp;=&amp;\\frac{\\var{conj(z3)^d1}}{\\var{abs(z3)^(2*d1)}}\\\\ &amp;=&amp;\\simplify[std]{{re(conj(z3^d1))}/{(abs(z3)^(2*d1))}+({im(conj(z3^d1))}/{round(abs(z3)^(2*d1))})*i} \\end{eqnarray*}\\]<br/>d)<br/>Es ist $i^2=-1,\\;\\;i^3=-i,\\;\\;i^4=1$.</p>\n<p>Also gilt f&uuml;r $n=4m+r,\\;\\;0\\le r\\le 3$, dass \\[i^n=i^{4m+r}=(i^4)^m \\cdot i^r=i^r\\]<br/>In diesem Fall ist $\\var{n}=4\\cdot\\var{m}+\\var{rem}$, also folgt<br/>\\[i^{\\var{n}}=i^{\\var{rem}}=\\simplify{i^{rem}}\\]</p>", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "extensions": [], "variables": {"s1": {"name": "s1", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "templateType": "anything"}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "random(2000..5000)", "description": "", "templateType": "anything"}, "rem": {"name": 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