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Lösung 3.2:6f

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Lösning 3.2:6f moved to Solution 3.2:6f: Robot: moved page)
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We can write every factor in the numerator and denominator in polar form and then use the arithmetical rules for multiplication and division in polar form:
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<center> [[Image:3_2_6f-1(2).gif]] </center>
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<math>\begin{array}{*{35}l}
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<center> [[Image:3_2_6f-2(2).gif]] </center>
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\bullet \quad r_{1}\left( \cos \alpha +i\sin \alpha \right)\centerdot r_{2}\left( \cos \beta +i\sin \beta \right)=r_{1}r_{2} \\
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\bullet \quad \frac{r_{1}\left( \cos \alpha +i\sin \alpha \right)}{r_{2}\left( \cos \beta +i\sin \beta \right)}=\frac{r_{1}}{r_{2}}\left( \cos \left( \alpha -\beta \right)+i\sin \left( \alpha -\beta \right) \right) \\
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\end{array}</math>
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In fact, most of the work consists of writing all the factors in polar form:
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[[Image:3_2_6_f1_bild.gif]][[Image:3_2_6_f1_bildtext.gif]]
[[Image:3_2_6_f1_bild.gif]][[Image:3_2_6_f1_bildtext.gif]]
[[Image:3_2_6_f2_bild.gif]][[Image:3_2_6_f2_bildtext.gif]]
[[Image:3_2_6_f2_bild.gif]][[Image:3_2_6_f2_bildtext.gif]]
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The whole expression becomes
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<math>\begin{align}
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& \frac{\left( 2+2i \right)\left( 1+i\sqrt{3} \right)}{3i\left( \sqrt{12}-2i \right)}=\frac{2\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)\centerdot 2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)}{3\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\centerdot 4\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right)} \\
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& =\frac{4\sqrt{2}\left( \cos \left( \frac{\pi }{4}+\frac{\pi }{3} \right)+i\sin \left( \frac{\pi }{4}+\frac{\pi }{3} \right) \right)}{12\left( \cos \left( \frac{\pi }{2}-\frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{2}-\frac{\pi }{6} \right) \right)} \\
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& =\frac{4\sqrt{2}\left( \cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12} \right)}{12\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)} \\
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& =\frac{4\sqrt{2}}{12}\left( \cos \left( \frac{7\pi }{12}-\frac{\pi }{3} \right)+i\sin \left( \frac{7\pi }{12}-\frac{\pi }{3} \right) \right) \\
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& =\frac{\sqrt{2}}{3}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
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\end{align}</math>

Version vom 11:00, 23. Okt. 2008

We can write every factor in the numerator and denominator in polar form and then use the arithmetical rules for multiplication and division in polar form:


r1cos+isinr2cos+isin=r1r2r2cos+isinr1cos+isin=r2r1cos−+isin−


In fact, most of the work consists of writing all the factors in polar form:


Image:3_2_6_f1_bild.gifImage:3_2_6_f1_bildtext.gif

Image:3_2_6_f2_bild.gifImage:3_2_6_f2_bildtext.gif

The whole expression becomes


3i12−2i2+2i1+i3=22cos4+isin42cos3+isin33cos2+isin24cos−6+isin−6=12cos2−6+isin2−642cos4+3+isin4+3=12cos3+isin342cos712+isin712=1242cos127−3+isin127−3=32cos4+isin4

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.2:6f“