Solution 3.3:1b

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Current revision (08:55, 30 October 2008) (edit) (undo)
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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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<center> [[Bild:3_3_1b.gif]] </center>
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<center>[[Image:3_3_1_b.gif]] [[Image:3_3_1_b_text.gif]]</center>
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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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and de Moivre's formula gives
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{{Displayed math||<math>\begin{align}
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\Bigl(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Bigr)^{12}
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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\cdot (\cos 4\pi + i\sin 4\pi)\\[5pt]
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&= 1\cdot (1+i\cdot 0)\\[5pt]
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&= 1\,\textrm{.}
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\end{align}</math>}}

Current revision

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}

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