Lösung 3.2:6f - Versionsgeschichte

Moni um 17:19, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 17:19:39 GMT

← Nächstältere Version		Version vom 17:19, 22. Aug. 2009
Zeile 1:		Zeile 1:
-	Schreiben wir den Zähler und Nenner in Polarform, ~~koönnen~~ wir die Divisions- und Multilpikationsregeln für komplexe Zahlen in Polarform verwenden:	+	Schreiben wir den Zähler und Nenner in Polarform, können wir die Divisions- und Multilpikationsregeln für komplexe Zahlen in Polarform verwenden:

	:*<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>		:*<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>

Martin um 15:05, 13. Mai 2009

Martin — Wed, 13 May 2009 15:05:24 GMT

← Nächstältere Version		Version vom 15:05, 13. Mai 2009
Zeile 1:		Zeile 1:
-	~~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~~:	+	Schreiben wir den Zähler und Nenner in Polarform, koönnen wir die Divisions- und Multilpikationsregeln für komplexe Zahlen in Polarform verwenden:

	:*<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>		:*<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>
Zeile 5:		Zeile 5:
	:*<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(\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>

-	~~In fact, most of the work consists of writing all the factors~~ in ~~polar form~~:	+	Wir schreiben die Zahlen also in Polarform:

	{\| align="center"		{\| align="center"
Zeile 15:		Zeile 15:
	\|}		\|}

-	~~The whole expression becomes~~	+	und erhalten den Ausdruck

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

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

Tekbot — Wed, 11 Mar 2009 10:42:15 GMT

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

← Nächstältere Version

Version vom 10:42, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:10:34 GMT

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

← Nächstältere Version		Version vom 13:10, 10. Mär. 2009
Zeile 17:		Zeile 17:
	The whole expression becomes		The whole expression becomes

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{(2+2i)(1+i\sqrt{3}\,)}{3i(\sqrt{12}-2i)}		\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{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]

Tek um 14:42, 29. Okt. 2008

Tek — Wed, 29 Oct 2008 14:42:40 GMT

← Nächstältere Version		Version vom 14:42, 29. Okt. 2008
Zeile 1:		Zeile 1:
	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:

		+	:*<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>

-	<math>~~\begin{array}{*{35}l}~~	+	:*<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>
-	~~\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} \\~~	+
-	~~\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~~) \\	+
-	\~~end~~{~~array~~}</math>	+
-		+

	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:

-		+	{\| align="center"
-	[[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]]
		+	\|}

	The whole expression becomes		The whole expression becomes

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{(2+2i)(1+i\sqrt{3}\,)}{3i(\sqrt{12}-2i)}
-	& \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~~)} \\	+	&= \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}\~~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~~)} \\	+	&= \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}\~~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~~)} \\	+	&= \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}\~~left~~( \cos \~~left~~( \frac{7\pi }{12}-\frac{\pi }{3} \~~right~~)+i\sin \~~left~~( \frac{7\pi }{12}-\frac{\pi }{3} \~~right~~) \~~right~~) \\	+	&= \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}\~~left~~( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \~~right~~) \\	+	&= \frac{\sqrt{2}}{3}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian um 11:00, 23. Okt. 2008

Ian — Thu, 23 Oct 2008 11:00:59 GMT

← Nächstältere Version		Version vom 11:00, 23. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	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:
-	~~<center> [[Image:3_2_6f-~~1(2)~~.gif]] </center>~~	+
-	{~~{NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\begin{array}{*{35}l}
-	~~<center> [[Image:3_2_6f-~~2(2)~~.gif]]~~ </~~center~~>	+	\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} \\
-	~~{{NAVCONTENT_STOP}}~~	+	\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) \\
		+	\end{array}</math>
		+
		+
		+	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_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]]
		+
		+	The whole expression becomes
		+
		+
		+	<math>\begin{align}
		+	& \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)} \\
		+	& =\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)} \\
		+	& =\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)} \\
		+	& =\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) \\
		+	& =\frac{\sqrt{2}}{3}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
		+	\end{align}</math>

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

Tekbot — Tue, 16 Sep 2008 15:02:41 GMT

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

← Nächstältere Version

Version vom 15:02, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:11:04 GMT

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

← Nächstältere Version		Version vom 08:11, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_6f-1(2).gif]] </center>	+	<center> [[Image:3_2_6f-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[~~Bild~~:3_2_6_f1_bild.gif]][[~~Bild~~:3_2_6_f1_bildtext.gif]]	+	[[Image:3_2_6_f1_bild.gif]][[Image:3_2_6_f1_bildtext.gif]]

-	[[~~Bild~~:3_2_6_f2_bild.gif]][[~~Bild~~:3_2_6_f2_bildtext.gif]]	+	[[Image:3_2_6_f2_bild.gif]][[Image:3_2_6_f2_bildtext.gif]]

Madde um 16:24, 19. Aug. 2008

Madde — Tue, 19 Aug 2008 16:24:47 GMT

← Nächstältere Version		Version vom 16:24, 19. Aug. 2008
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Oskar — Mon, 07 Apr 2008 12:20:12 GMT

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