Lösung 3.2:6d

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We can determine the number's magnitude directly using the distance formula
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<center> [[Image:3_2_6d.gif]] </center>
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<math>\begin{align}
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& \left| \sqrt{10}+\sqrt{30}i \right|=\sqrt{\left( \sqrt{10} \right)^{2}+\left( \sqrt{30} \right)^{2}}=\sqrt{10+30} \\
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& =\sqrt{40}=\sqrt{4\centerdot 10}=2\sqrt{10} \\
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\end{align}</math>
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In addition, the number lies in the first quadrant and we can therefore determine the argument using simple trigonometry.
[[Image:3_2_6_d_bild.gif]] [[Image:3_2_6_d_bildtext.gif]]
[[Image:3_2_6_d_bild.gif]] [[Image:3_2_6_d_bildtext.gif]]
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The polar form is
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<math>2\sqrt{10}\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)</math>

Version vom 10:30, 23. Okt. 2008

We can determine the number's magnitude directly using the distance formula


\displaystyle \begin{align} & \left| \sqrt{10}+\sqrt{30}i \right|=\sqrt{\left( \sqrt{10} \right)^{2}+\left( \sqrt{30} \right)^{2}}=\sqrt{10+30} \\ & =\sqrt{40}=\sqrt{4\centerdot 10}=2\sqrt{10} \\ \end{align}


In addition, the number lies in the first quadrant and we can therefore determine the argument using simple trigonometry.


Image:3_2_6_d_bild.gif Image:3_2_6_d_bildtext.gif


The polar form is


\displaystyle 2\sqrt{10}\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:6d“