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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<center> [[Image:3_3_1a-1(2).gif]] </center>
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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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<center> [[Image:3_3_1a-2(2).gif]] </center>
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The equation above is called de Moivre's formula.
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The plan is therefore to rewrite
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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]]
[[Image:3_3_1_a1.gif]] [[Image:3_3_1_a_text.gif]]
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Using the calculations above, we see that
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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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<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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& =2^{\frac{1}{2}\centerdot 12}\left( \cos 3\pi +i\sin 3\pi \right) \\
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& =2^{6}\left( -1+i\centerdot 0 \right)=64\centerdot \left( -1 \right)=-64 \\
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& \\
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\end{align}</math>

Version vom 07:06, 24. 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 \left( r\left( \cos \alpha +i\sin \alpha \right) \right)^{n}=r^{n}\left( \cos n\alpha +i\sin n\alpha \right)


The equation above is called de Moivre's formula.

The plan is therefore to rewrite \displaystyle \text{1}+i in polar form, raised the expression to the power \displaystyle \text{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 \text{1}+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)


de Moivre's formula now gives


\displaystyle \begin{align} & \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) \\ & =2^{\frac{1}{2}\centerdot 12}\left( \cos 3\pi +i\sin 3\pi \right) \\ & =2^{6}\left( -1+i\centerdot 0 \right)=64\centerdot \left( -1 \right)=-64 \\ & \\ \end{align}

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