Lösung 3.3:1b

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{{NAVCONTENT_START}}
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First, we write the number
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<center> [[Image:3_3_1b.gif]] </center>
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<math>\frac{1}{2}+i\frac{\sqrt{3}}{2}</math>
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{{NAVCONTENT_STOP}}
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in polar form.
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[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]
[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]
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Thus,
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<math>\frac{1}{2}+i\frac{\sqrt{3}}{2}=1\centerdot \left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)</math>
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and de Moivre's formula gives
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<math>\begin{align}
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& \left( \frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{12}=1^{12}\centerdot \left( \cos 12\centerdot \frac{\pi }{3}+i\sin 12\centerdot \frac{\pi }{3} \right) \\
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& =1\centerdot \left( \cos 4\pi +i\sin 4\pi \right) \\
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& =1\centerdot \left( 1+i\centerdot 0 \right)=1 \\
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\end{align}</math>

Version vom 07:14, 24. Okt. 2008

First, we write the number \displaystyle \frac{1}{2}+i\frac{\sqrt{3}}{2} in polar form.


Image:3_3_1_b.gif Image:3_3_1_b_text.gif

Thus,


\displaystyle \frac{1}{2}+i\frac{\sqrt{3}}{2}=1\centerdot \left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)


and de Moivre's formula gives


\displaystyle \begin{align} & \left( \frac{1}{2}+i\frac{\sqrt{3}}{2} \right)^{12}=1^{12}\centerdot \left( \cos 12\centerdot \frac{\pi }{3}+i\sin 12\centerdot \frac{\pi }{3} \right) \\ & =1\centerdot \left( \cos 4\pi +i\sin 4\pi \right) \\ & =1\centerdot \left( 1+i\centerdot 0 \right)=1 \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:1b“