// Numbas version: finer_feedback_settings
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anwenden k&ouml;nnen und dann eine reelle Zahl im Nenner erhalten.<br/>F&uuml;r \\[ z = \\frac{a+bi}{c+di},\\;\\; c+di \\neq 0\\] haben wir<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*} \\]</p>\n<p>(Statt der Formel sollten Sie sich die Methode merken.)<br/><br/><strong>(a)</strong><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/><strong>(b)</strong><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/><strong>(c)</strong><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/><strong>(d)</strong><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", "simplifyfractions"]}, "extensions": [], "variables": {"s1": {"name": "s1", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", 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