Lösung 3.2:6f

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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:
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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:*<math>r_1(\cos\alpha + i\sin\alpha)\cdot r_2(\cos\beta + i\sin\beta) = r_1r_2\bigl(\cos(\alpha+\beta)+i\sin(\alpha+\beta)\bigr)\,,</math>
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<math>\begin{array}{*{35}l}
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:*<math>\frac{r_1(\cos\alpha + i\sin\alpha)}{r_2(\cos\beta + i\sin\beta)} = \frac{r_1}{r_2}\bigl(\cos(\alpha-\beta) + i\sin (\alpha-\beta)\bigr)\,\textrm{.}</math>
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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:
In fact, most of the work consists of writing all the factors in polar form:
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{| align="center"
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[[Image:3_2_6_f1_bild.gif]][[Image:3_2_6_f1_bildtext.gif]]
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||[[Image:3_2_6_f1_bild.gif]]
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||[[Image:3_2_6_f1_bildtext.gif]]
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[[Image:3_2_6_f2_bild.gif]][[Image:3_2_6_f2_bildtext.gif]]
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|-
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||[[Image:3_2_6_f2_bild.gif]]
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||[[Image:3_2_6_f2_bildtext.gif]]
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|}
The whole expression becomes
The whole expression becomes
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{(2+2i)(1+i\sqrt{3}\,)}{3i(\sqrt{12}-2i)}
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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{2\sqrt{2}\Bigl(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\Bigr)\cdot 2\Bigl( \cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\Bigr)}{3\Bigl(\cos\dfrac{\pi}{2}+i\sin \dfrac{\pi}{2}\Bigr)\cdot 4\Bigl(\cos\Bigl(-\dfrac{\pi}{6}\Bigr)+i\sin\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)}\\[5pt]
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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}\Bigl(\cos\Bigl(\dfrac{\pi}{4}+\dfrac{\pi}{3}\Bigr) + i\sin\Bigl( \dfrac{\pi}{4}+\dfrac{\pi}{3}\Bigr)\Bigr)}{12\Bigl(\cos\Bigl(\dfrac{\pi}{2}-\dfrac{\pi}{6}\Bigr)+i\sin\Bigl(\dfrac{\pi}{2}-\dfrac{\pi}{6}\Bigr)\Bigr)}\\[5pt]
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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}\Bigl(\cos\dfrac{7\pi}{12}+i\sin\dfrac{7\pi}{12}\Bigr)}{12\Bigl(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\Bigr)}\\[5pt]
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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{4\sqrt{2}}{12}\Bigl(\cos\Bigl(\frac{7\pi}{12}-\frac{\pi}{3}\Bigr) + i\sin \Bigl(\frac{7\pi}{12}-\frac{\pi}{3}\Bigr)\Bigr)\\[5pt]
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& =\frac{\sqrt{2}}{3}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
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&= \frac{\sqrt{2}}{3}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 14:42, 29. 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:

  • \displaystyle r_1(\cos\alpha + i\sin\alpha)\cdot r_2(\cos\beta + i\sin\beta) = r_1r_2\bigl(\cos(\alpha+\beta)+i\sin(\alpha+\beta)\bigr)\,,
  • \displaystyle \frac{r_1(\cos\alpha + i\sin\alpha)}{r_2(\cos\beta + i\sin\beta)} = \frac{r_1}{r_2}\bigl(\cos(\alpha-\beta) + i\sin (\alpha-\beta)\bigr)\,\textrm{.}

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

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

The whole expression becomes

\displaystyle \begin{align}

\frac{(2+2i)(1+i\sqrt{3}\,)}{3i(\sqrt{12}-2i)} &= \frac{2\sqrt{2}\Bigl(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\Bigr)\cdot 2\Bigl( \cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\Bigr)}{3\Bigl(\cos\dfrac{\pi}{2}+i\sin \dfrac{\pi}{2}\Bigr)\cdot 4\Bigl(\cos\Bigl(-\dfrac{\pi}{6}\Bigr)+i\sin\Bigl(-\dfrac{\pi}{6}\Bigr)\Bigr)}\\[5pt] &= \frac{4\sqrt{2}\Bigl(\cos\Bigl(\dfrac{\pi}{4}+\dfrac{\pi}{3}\Bigr) + i\sin\Bigl( \dfrac{\pi}{4}+\dfrac{\pi}{3}\Bigr)\Bigr)}{12\Bigl(\cos\Bigl(\dfrac{\pi}{2}-\dfrac{\pi}{6}\Bigr)+i\sin\Bigl(\dfrac{\pi}{2}-\dfrac{\pi}{6}\Bigr)\Bigr)}\\[5pt] &= \frac{4\sqrt{2}\Bigl(\cos\dfrac{7\pi}{12}+i\sin\dfrac{7\pi}{12}\Bigr)}{12\Bigl(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\Bigr)}\\[5pt] &= \frac{4\sqrt{2}}{12}\Bigl(\cos\Bigl(\frac{7\pi}{12}-\frac{\pi}{3}\Bigr) + i\sin \Bigl(\frac{7\pi}{12}-\frac{\pi}{3}\Bigr)\Bigr)\\[5pt] &= \frac{\sqrt{2}}{3}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)\,\textrm{.} \end{align}

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