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Lösung 3.3:1e

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.3:1e moved to Solution 3.3:1e: Robot: moved page)
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One part of the quotient has rather high exponents and this indicates that we ought to use polar form for the calculation.
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<center> [[Image:3_3_1e-1(2).gif]] </center>
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First, we write
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<math>1+i\sqrt{3}</math>,
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<center> [[Image:3_3_1e-2(2).gif]] </center>
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<math>\text{1}-i</math>
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and
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<math>\sqrt{3}-i</math>
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in polar form.
[[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]]
[[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]]
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This shows that
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<math>\begin{align}
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& 1+i\sqrt{3}=2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\
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& \text{1}-i=\sqrt{2}\left( \cos \left( -\frac{\pi }{4} \right)+i\sin \left( -\frac{\pi }{4} \right) \right) \\
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& \sqrt{3}-i=2\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \\
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\end{align}</math>
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Now, with the help of de Moivre's formula,
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<math>\begin{align}
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& \frac{\left( 1+i\sqrt{3} \right)\left( \text{1}-i \right)^{8}}{\left( \sqrt{3}-i \right)^{9}}=\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\left( \sqrt{2}\left( \cos \left( -\frac{\pi }{4} \right)+i\sin \left( -\frac{\pi }{4} \right) \right) \right)^{8}}{\left( 2\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \right)^{9}} \\
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& =\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\left( \sqrt{2} \right)^{8}\left( \cos \left( 8\centerdot \left( -\frac{\pi }{4} \right) \right)+i\sin \left( 8\centerdot \left( -\frac{\pi }{4} \right) \right) \right)}{2^{9}\left( \cos \left( 9\centerdot \left( -\frac{\pi }{6} \right) \right)+i\sin \left( 9\centerdot \left( -\frac{\pi }{6} \right) \right) \right)} \\
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& =\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\centerdot 2^{\frac{1}{2}\centerdot 8}\left( \cos \left( -2\pi \right)+i\sin \left( -2\pi \right) \right)}{2^{9}\left( \cos \left( -\frac{3\pi }{2} \right)+i\sin \left( -\frac{3\pi }{2} \right) \right)} \\
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& =\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)\centerdot 2^{4}\left( 1+i\centerdot 0 \right)}{2^{9}\left( \cos \left( -\frac{3\pi }{2} \right)+i\sin \left( -\frac{3\pi }{2} \right) \right)} \\
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& =\frac{2\centerdot 2^{4}}{2^{9}}\left( \cos \left( \frac{\pi }{3}-\left( -\frac{3\pi }{2} \right) \right)+i\sin \left( \frac{\pi }{3}-\left( -\frac{3\pi }{2} \right) \right) \right) \\
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& =\frac{2^{5}}{2^{9}}\left( \cos \left( \frac{\pi }{3}+\frac{3\pi }{2} \right)+i\sin \left( \frac{\pi }{3}+\frac{3\pi }{2} \right) \right) \\
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& =\frac{1}{2^{4}}\left( \cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6} \right) \\
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& =\frac{1}{16}\left( \cos \frac{12\pi -\pi }{6}+i\sin \frac{12\pi -\pi }{6} \right) \\
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& =\frac{1}{16}\left( \cos \left( 2\pi -\frac{\pi }{6} \right)+i\sin \left( 2\pi -\frac{\pi }{6} \right) \right) \\
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& =\frac{1}{16}\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \\
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& =\frac{1}{16}\left( \frac{\sqrt{3}}{2}-\frac{i}{2} \right)=\frac{1}{32}\left( \sqrt{3}-i \right) \\
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\end{align}</math>

Version vom 08:21, 24. Okt. 2008

One part of the quotient has rather high exponents and this indicates that we ought to use polar form for the calculation.

First, we write 1+i3 , 1−i and 3−i  in polar form.


Image:3_3_1_e.gif Image:3_3_1_e_text.gif

This shows that


1+i3=2cos3+isin31−i=2cos−4+isin−43−i=2cos−6+isin−6


Now, with the help of de Moivre's formula,


3−i91+i31−i8=2cos−6+isin−692cos3+isin32cos−4+isin−48=29cos9−6+isin9−62cos3+isin328cos8−4+isin8−4=29cos−23+isin−232cos3+isin32218cos−2+isin−2=29cos−23+isin−232cos3+isin3241+i0=29224cos3−−23+isin3−−23=2925cos3+23+isin3+23=124cos611+isin611=116cos612−+isin612−=116cos2−6+isin2−6=116cos−6+isin−6=11623−i2=1323−i

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