Lösung 3.2:4d

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For magnitudes of quotients, we have the arithmetical rule
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<center> [[Image:3_2_4d.gif]] </center>
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<math>\left| \frac{z}{w} \right|=\frac{\left| z \right|}{\left| w \right|}</math>
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We can therefore take the magnitude of the numerator and denominator separately and then divide the magnitudes by each other:
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<math>\begin{align}
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& \left| \frac{3-4i}{3+2i} \right|=\frac{\left| 3-4i \right|}{\left| 3+2i \right|}=\frac{\sqrt{3^{2}+\left( -4 \right)^{2}}}{\sqrt{3^{2}+2^{2}}}=\frac{\sqrt{9+16}}{\sqrt{9+4}} \\
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& =\frac{\sqrt{25}}{\sqrt{13}}=\frac{5}{\sqrt{13}} \\
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\end{align}</math>

Version vom 15:39, 22. Okt. 2008

For magnitudes of quotients, we have the arithmetical rule


\displaystyle \left| \frac{z}{w} \right|=\frac{\left| z \right|}{\left| w \right|}


We can therefore take the magnitude of the numerator and denominator separately and then divide the magnitudes by each other:


\displaystyle \begin{align} & \left| \frac{3-4i}{3+2i} \right|=\frac{\left| 3-4i \right|}{\left| 3+2i \right|}=\frac{\sqrt{3^{2}+\left( -4 \right)^{2}}}{\sqrt{3^{2}+2^{2}}}=\frac{\sqrt{9+16}}{\sqrt{9+4}} \\ & =\frac{\sqrt{25}}{\sqrt{13}}=\frac{5}{\sqrt{13}} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:4d“