// Numbas version: exam_results_page_options {"name": "Rechnen mit komplexen Zahlen III", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Rechnen mit komplexen Zahlen III", "tags": [], "metadata": {"description": "
Inverse and division of complex numbers. Four parts.
\nOriginal: https://numbas.mathcentre.ac.uk/question/11786/arithmetics-of-complex-numbers-iii/ by Newcastle University Mathematics and Statistics
\nTranslated to German.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Schreiben Sie die folgenden Ausdrücke in der Form $a+bi$ mit $a,b\\in\\mathbb R$.
\nGeben Sie $a$ und $b$ als Bruchzahlen oder als ganze Zahlen an.
", "advice": "Um einen Bruch komplexer Zahlen in der Form $a+bi$ zu schreiben, erweitern wir so, dass wir die dritte binomische Formel anwenden können und dann eine reelle Zahl im Nenner erhalten.
Für \\[ z = \\frac{a+bi}{c+di},\\;\\; c+di \\neq 0\\] haben wir
\\[\\begin{eqnarray*} z&=&\\frac{a+bi}{c+di}\\\\ &=&\\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\\\ &=&\\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\\\\ &=&\\frac{ac+bd}{c^2+d^2}+\\frac{bc-ad}{c^2+d^2}i \\end{eqnarray*} \\]
(Statt der Formel sollten Sie sich die Methode merken.)
(a)
\\[\\begin{eqnarray*}\\simplify[std]{{c1}/{z1}} &=&\\simplify[std]{({c1}*{conj(z1)})/({z1}*{conj(z1)})}\\\\ &=&\\simplify[std]{{c1*conj(z1)}/{abs(z1)^2}}\\\\ &=& \\simplify[std]{{c1*re(z1)}/{abs(z1)^2}-{c1*im(z1)}/{abs(z1)^2}*i} \\end{eqnarray*} \\]
(b)
\\[\\begin{eqnarray*}\\simplify[std]{{c2}/{z2}} &=&\\simplify[std]{({c2}*{conj(z2)})/({z2}*{conj(z2)})}\\\\ &=&\\simplify[std]{{c2*conj(z2)}/{abs(z2)^2}}\\\\ &=& \\simplify[std]{{c2*re(z2)}/{abs(z2)^2}-{c2*im(z2)}/{abs(z2)^2}*i} \\end{eqnarray*} \\]
(c)
\\[\\begin{eqnarray*}\\simplify[std]{{z1}/{z3}} &=&\\simplify[std]{({z1}*{conj(z3)})/({z3}*{conj(z3)})}\\\\ &=&\\simplify[std]{{z1*conj(z3)}/{abs(z3)^2}}\\\\ &=& \\simplify[std]{{re(z1*conj(z3))}/{abs(z3)^2}+{im(z1*conj(z3))}/{abs(z3)^2}*i} \\end{eqnarray*} \\]
(d)
\\[\\begin{eqnarray*}\\simplify[std]{{z3}/{z2}} &=&\\simplify[std]{({z3}*{conj(z2)})/({z2}*{conj(z2)})}\\\\ &=&\\simplify[std]{{z3*conj(z2)}/{abs(z2)^2}}\\\\ &=& \\simplify[std]{{re(z3*conj(z2))}/{abs(z2)^2}+{im(z3*conj(z2))}/{abs(z2)^2}*i} \\end{eqnarray*} \\]
$\\displaystyle \\simplify[std]{{c1}/{z1}} = $ [[0]]
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\nSchreiben Sie Ihre Antwort ohne Klammern.
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