Lösung 3.2:6e - Versionsgeschichte

Moni um 16:42, 22. Aug. 2009

2009-08-22T16:42:23Z

← Nächstältere Version		Version vom 16:42, 22. Aug. 2009
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-	~~Können~~ wir den Zähler und den Nenner auf Polarform bringen, können wir die Division einfach ausführen, indem wir die Betrage dividieren, und die Argumente subtrahieren.	+	Wenn wir den Zähler und den Nenner auf Polarform bringen, können wir die Division einfach ausführen, indem wir die Betrage dividieren und die Argumente subtrahieren.

	{{Abgesetzte Formel\|\|<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>}}		{{Abgesetzte Formel\|\|<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>}}

-	Daher schreiben wir den Zähler und den ~~nenner auf~~ Polarform:	+	Daher schreiben wir den Zähler und den Nenner in Polarform:

	[[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]]

Martin um 14:57, 13. Mai 2009

2009-05-13T14:57:10Z

← Nächstältere Version		Version vom 14:57, 13. Mai 2009
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-	Können wir den Zähler und den Nenner auf Polarform bringen, können wir die Division einfach ausführen, indem wir	+	Können wir den Zähler und den Nenner auf Polarform bringen, können wir die Division einfach ausführen, indem wir die Betrage dividieren, und die Argumente subtrahieren.
-	~~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,~~	+

	{{Abgesetzte Formel\|\|<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>}}		{{Abgesetzte Formel\|\|<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>}}

-	~~Thus, we determine the numerator's and denominator's polar forms~~:	+	Daher schreiben wir den Zähler und den nenner auf Polarform:

	[[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]]

-	~~Now, we obtain~~	+	und erhalten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Martin um 13:13, 13. Mai 2009

2009-05-13T13:13:35Z

← Nächstältere Version		Version vom 13:13, 13. Mai 2009
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		+	Können wir den Zähler und den Nenner auf Polarform bringen, können wir die Division einfach ausführen, indem wir
	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,

Tekbot: Solution 3.2:6e moved to Lösung 3.2:6e: Robot: moved page

2009-03-11T10:41:55Z

Solution 3.2:6e moved to Lösung 3.2:6e: Robot: moved page

← Nächstältere Version

Version vom 10:41, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

2009-03-10T13:10:24Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 13:10, 10. Mär. 2009
Zeile 1:		Zeile 1:
	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,

-	{{~~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>}}	+	{{Abgesetzte Formel\|\|<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>}}

	Thus, we determine the numerator's and denominator's polar forms:		Thus, we determine the numerator's and denominator's polar forms:
Zeile 9:		Zeile 9:
	Now, we obtain		Now, we obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{1+i\sqrt{3}}{1+i}		\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\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]

Tek um 14:22, 29. Okt. 2008

2008-10-29T14:22:05Z

← Nächstältere Version		Version vom 14:22, 29. Okt. 2008
Zeile 1:		Zeile 1:
	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,

-		+	{{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>}}
-	<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>.	+

	Thus, we determine the numerator's and denominator's polar forms:		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]]		[[Image:3_2_6_e_bild.gif]] [[Image:3_2_6_e_bildtext.gif]]
-

	Now, we obtain		Now, we obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{1+i\sqrt{3}}{1+i}
-	& \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~~)} \\	+	&= \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}}\~~left~~( \cos \~~left~~( \frac{\pi }{3}-\frac{\pi }{4} \~~right~~)+i\sin \~~left~~( \frac{\pi }{3}-\frac{\pi }{4} \~~right~~) \~~right~~) \\	+	&= \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}\~~left~~( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \~~right~~) \\	+	&= \sqrt{2}\Bigl(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\Bigr)\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian um 10:38, 23. Okt. 2008

2008-10-23T10:38:38Z

← Nächstältere Version		Version vom 10:38, 23. Okt. 2008
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-	~~{{NAVCONTENT_START}}~~	+	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,
-	<~~center~~> ~~[[Image:3_2_6e-~~1(2)~~.gif]] </center>~~	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<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>.
-	~~<center> [[Image:3_2_6e~~-2(2)~~.gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+	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]]		[[Image:3_2_6_e_bild.gif]] [[Image:3_2_6_e_bildtext.gif]]
		+
		+
		+	Now, we obtain
		+
		+
		+	<math>\begin{align}
		+	& \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)} \\
		+	& =\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) \\
		+	& =\sqrt{2}\left( \cos \frac{\pi }{12}+i\sin \frac{\pi }{12} \right) \\
		+	\end{align}</math>

Tekbot: Lösning 3.2:6e moved to Solution 3.2:6e: Robot: moved page

2008-09-16T15:02:21Z

Lösning 3.2:6e moved to Solution 3.2:6e: Robot: moved page

← Nächstältere Version

Version vom 15:02, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T08:10:54Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Madde um 16:20, 19. Aug. 2008

2008-08-19T16:20:19Z

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