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Composite multiplication and division of complex numbers. Two parts.

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Original: https://numbas.mathcentre.ac.uk/question/11787/arithmetics-of-complex-numbers-iv/ by Newcastle University Mathematics and Statistics

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Translated to German

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Drücken Sie die unten gegebenen komplexen Zahlen $z$ in der Form $a+bi$ aus.

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Geben Sie $a$ und $b$ als Bruchzahlen oder als ganze Zahlen an.

", "advice": "

a)
\\[\\begin{eqnarray*}z=\\simplify[!collectNumbers]{({z3}*{z2})/{z1}} &=&\\simplify[!collectNumbers]{({z3}*{z2}*{conj(z1)})/({z1}*{conj(z1)})}\\\\ &=&\\simplify[!collectNumbers]{({z3*z2}*{conj(z1)})/({abs(z1)^2})}\\\\ &=&\\simplify[!collectNumbers]{{z3*z2*conj(z1)}/{abs(z1)^2}}\\\\ &=& \\simplify[std]{{re(z3*z2*conj(z1))}/{abs(z1)^2}+{im(z3*z2*conj(z1))}/{abs(z1)^2}*i} \\end{eqnarray*} \\]

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b)
\\[\\begin{eqnarray*}z= \\simplify[!collectNumbers]{({z2}*{z1})}(\\var{z3})^{-1} &=& \\simplify[!collectNumbers]{({z2}*{z1})/{z3}}\\\\ &=&\\simplify[!collectNumbers]{({z2}*{z1}*{conj(z3)})/({z3}*{conj(z3)})}\\\\ &=&\\simplify[!collectNumbers]{({z2*z1}*{conj(z3)})/({abs(z3)^2})}\\\\ &=&\\simplify[!collectNumbers]{{z2*z1*conj(z3)}/{abs(z3)^2}}\\\\ &=& \\simplify[std]{{re(z2*z1*conj(z3))}/{abs(z3)^2}+{im(z2*z1*conj(z3))}/{abs(z3)^2}*i} \\end{eqnarray*} \\]

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\\[\\displaystyle z=\\simplify[!collectNumbers]{({z3}*{z2})/{z1}}\\]

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$z=\\;\\;$[[0]].

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\\[\\displaystyle z=\\simplify[!collectNumbers]{({z2}*{z1})}(\\var{z3})^{-1}\\]

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$z=\\;\\;$[[0]].

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