Lösung 3.2:6d

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We can determine the number's magnitude directly using the distance formula
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We can determine the number's magnitude directly using the distance formula,
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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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{{Displayed math||<math>\begin{align}
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\bigl|\sqrt{10}+\sqrt{30}i\bigr|
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&= \sqrt{\bigl(\sqrt{10}\,\bigr)^2+\bigl(\sqrt{30}\,\bigr)^2}\\[5pt]
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&= \sqrt{10+30}\\[5pt]
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&= \sqrt{40}\\[5pt]
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&= \sqrt{4\cdot 10}\\[5pt]
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&= 2\sqrt{10}\,\textrm{.}
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\end{align}</math>}}
In addition, the number lies in the first quadrant and we can therefore determine the argument using simple trigonometry.
In addition, the number lies in the first quadrant and we can therefore determine the argument using simple trigonometry.
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[[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
The polar form is
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{{Displayed math||<math>2\sqrt{10}\Bigl( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \Bigr)\,\textrm{.}</math>}}
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<math>2\sqrt{10}\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)</math>
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Version vom 13:56, 29. Okt. 2008

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

\displaystyle \begin{align}

\bigl|\sqrt{10}+\sqrt{30}i\bigr| &= \sqrt{\bigl(\sqrt{10}\,\bigr)^2+\bigl(\sqrt{30}\,\bigr)^2}\\[5pt] &= \sqrt{10+30}\\[5pt] &= \sqrt{40}\\[5pt] &= \sqrt{4\cdot 10}\\[5pt] &= 2\sqrt{10}\,\textrm{.} \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}\Bigl( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \Bigr)\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:6d“