Lösung 3.3:1b

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First, we write the number
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First, we write the number <math>\frac{1}{2}+i\frac{\sqrt{3}}{2}</math> in polar form.
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<math>\frac{1}{2}+i\frac{\sqrt{3}}{2}</math>
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in polar form.
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<center>[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]</center>
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[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]
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Thus,
Thus,
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{{Displayed math||<math>\frac{1}{2}+i\frac{\sqrt{3}}{2} = 1\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)</math>}}
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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
and de Moivre's formula gives
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)^{12}
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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^{12}\cdot\Bigl(\cos\Bigl(12\cdot\frac{\pi}{3}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{3}\Bigr)\Bigr)\\[5pt]
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& =1\centerdot \left( \cos 4\pi +i\sin 4\pi \right) \\
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&= 1\cdot (\cos 4\pi + i\sin 4\pi)\\[5pt]
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& =1\centerdot \left( 1+i\centerdot 0 \right)=1 \\
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&= 1\cdot (1+i\cdot 0)\\[5pt]
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\end{align}</math>
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&= 1\,\textrm{.}
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\end{align}</math>}}

Version vom 08:55, 30. 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\cdot \Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)

and de Moivre's formula gives

\displaystyle \begin{align}

\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)^{12} &= 1^{12}\cdot\Bigl(\cos\Bigl(12\cdot\frac{\pi}{3}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{3}\Bigr)\Bigr)\\[5pt] &= 1\cdot (\cos 4\pi + i\sin 4\pi)\\[5pt] &= 1\cdot (1+i\cdot 0)\\[5pt] &= 1\,\textrm{.} \end{align}

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