Lösung 3.3:1a

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Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form:
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Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form,
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<math>\left( r\left( \cos \alpha +i\sin \alpha \right) \right)^{n}=r^{n}\left( \cos n\alpha +i\sin n\alpha \right)</math>
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{{Displayed math||<math>\bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}</math>}}
The equation above is called de Moivre's formula.
The equation above is called de Moivre's formula.
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The plan is therefore to rewrite
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The plan is therefore to rewrite <math>1+i</math> in polar form, raise the expression to the power <math>12</math> using de Moivre's formula and then to write the answer in the form <math>a+ib</math>.
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<math>\text{1}+i</math>
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in polar form, raised the expression to the power
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<math>\text{12}</math>
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using de Moivre's formula and then to write the answer in the form
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<math>a+ib</math>.
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[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]
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<center>[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]</center>
Using the calculations above, we see that
Using the calculations above, we see that
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{{Displayed math||<math>1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}</math>}}
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<math>\text{1}+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)</math>
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De Moivre's formula now gives
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de Moivre's formula now gives
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \left( \text{1}+i \right)^{12}=\left( \sqrt{2} \right)^{12}\left( \cos 12\centerdot \frac{\pi }{4}+i\sin 12\centerdot \frac{\pi }{4} \right) \\
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(1+i)^{12}
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& =2^{\frac{1}{2}\centerdot 12}\left( \cos 3\pi +i\sin 3\pi \right) \\
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&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos \Bigl(12\cdot\frac{\pi}{4}\Bigr) + i\sin \Bigl(12\cdot\frac{\pi}{4}\Bigr)\Bigr)\\[5pt]
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& =2^{6}\left( -1+i\centerdot 0 \right)=64\centerdot \left( -1 \right)=-64 \\
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&= 2^{(1/2)\cdot 12}\Bigl(\cos 3\pi + i\sin 3\pi\Bigr)\\[5pt]
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& \\
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&= 2^6(-1+i\cdot 0)\\[5pt]
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\end{align}</math>
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&= 64\cdot (-1)\\[5pt]
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&= -64\,\textrm{.}
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\end{align}</math>}}

Version vom 08:50, 30. Okt. 2008

Powers are repeated multiplications and because multiplication is a relatively simple arithmetical operation when it is carried out in polar form, calculating powers also becomes fairly simple in polar form,

\displaystyle \bigl(r(\cos\alpha + i\sin\alpha)\bigr)^n = r^n(\cos n\alpha + i\sin n\alpha)\,\textrm{.}

The equation above is called de Moivre's formula.

The plan is therefore to rewrite \displaystyle 1+i in polar form, raise the expression to the power \displaystyle 12 using de Moivre's formula and then to write the answer in the form \displaystyle a+ib.

Image:3_3_1_a1.gif Image:3_3_1_a_text.gif

Using the calculations above, we see that

\displaystyle 1+i = \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \Bigr)\,\textrm{.}

De Moivre's formula now gives

\displaystyle \begin{align}

(1+i)^{12} &= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos \Bigl(12\cdot\frac{\pi}{4}\Bigr) + i\sin \Bigl(12\cdot\frac{\pi}{4}\Bigr)\Bigr)\\[5pt] &= 2^{(1/2)\cdot 12}\Bigl(\cos 3\pi + i\sin 3\pi\Bigr)\\[5pt] &= 2^6(-1+i\cdot 0)\\[5pt] &= 64\cdot (-1)\\[5pt] &= -64\,\textrm{.} \end{align}

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