Lösung 3.2:6e

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If we manage to write the numerator and denominator in polar form, we can then use the fact that when division is carried out in polar form it is particularly simple. Division means, namely, that the numerator's magnitude is divided by the denominator's magnitude and that the numerator's argument is subtracted from denominator's argument or, in written form,
If we manage to write the numerator and denominator in polar form, we can then use the fact that when division is carried out in polar form it is particularly simple. Division means, namely, that the numerator's magnitude is divided by the denominator's magnitude and that the numerator's argument is subtracted from denominator's argument or, in written form,
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{{Displayed math||<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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<math>\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)</math>.
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Thus, we determine the numerator's and denominator's polar forms:
Thus, we determine the numerator's and denominator's polar forms:
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[[Image:3_2_6_e_bild.gif]] [[Image:3_2_6_e_bildtext.gif]]
[[Image:3_2_6_e_bild.gif]] [[Image:3_2_6_e_bildtext.gif]]
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Now, we obtain
Now, we obtain
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{1+i\sqrt{3}}{1+i}
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& \frac{1+i\sqrt{3}}{1+i}=\frac{2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)}{\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)} \\
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&= \frac{2\Bigl(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\Bigr)}{\sqrt{2}\Bigl(\cos \dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\Bigr)}\\[5pt]
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& =\frac{2}{\sqrt{2}}\left( \cos \left( \frac{\pi }{3}-\frac{\pi }{4} \right)+i\sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) \right) \\
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&= \frac{2}{\sqrt{2}}\Bigl( \cos\Bigl(\frac{\pi}{3}-\frac{\pi}{4}\Bigr) + i\sin\Bigl( \frac{\pi}{3}-\frac{\pi}{4}\Bigr)\Bigr)\\[5pt]
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& =\sqrt{2}\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right) \\
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&= \sqrt{2}\Bigl(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\Bigr)\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}

Version vom 14:22, 29. Okt. 2008

If we manage to write the numerator and denominator in polar form, we can then use the fact that when division is carried out in polar form it is particularly simple. Division means, namely, that the numerator's magnitude is divided by the denominator's magnitude and that the numerator's argument is subtracted from denominator's argument or, in written form,

\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{.}

Thus, we determine the numerator's and denominator's polar forms:

Image:3_2_6_e_bild.gif Image:3_2_6_e_bildtext.gif

Now, we obtain

\displaystyle \begin{align}

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

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