Lösung 3.3:1d - Versionsgeschichte

Sekretariat1 um 07:50, 1. Sep. 2009

2009-09-01T07:50:40Z

← Nächstältere Version		Version vom 07:50, 1. Sep. 2009
Zeile 1:		Zeile 1:
	Wir bringen zuerst		Wir bringen zuerst
-	<math>1+i\sqrt{3}</math> und <math>\text{1}+i</math> ~~auf~~ Polarform und rechnen weiterhin in Polarform.	+	<math>1+i\sqrt{3}</math> und <math>\text{1}+i</math> in Polarform und rechnen weiterhin in Polarform.

	<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>		<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>
Zeile 11:		Zeile 11:
	\end{align}</math>}}		\end{align}</math>}}

-	und ~~also~~	+	und daraus folgt

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

Moni um 21:02, 22. Aug. 2009

2009-08-22T21:02:11Z

← Nächstältere Version		Version vom 21:02, 22. Aug. 2009
Zeile 1:		Zeile 1:
	Wir bringen zuerst		Wir bringen zuerst
-	<math>1+i\sqrt{3}</math> und <math>\text{1}+i</math> auf Polarform, und rechnen weiterhin in Polarform.	+	<math>1+i\sqrt{3}</math> und <math>\text{1}+i</math> auf Polarform und rechnen weiterhin in Polarform.

	<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>		<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>

Martin um 14:47, 18. Mai 2009

2009-05-18T14:47:02Z

← Nächstältere Version		Version vom 14:47, 18. Mai 2009
Zeile 1:		Zeile 1:
-	Because we are going to raise something to the power 12, the base in the expression should be written in polar form. In turn, the base consists of a quotient which is advantageous to calculate in polar form. Thus, it seems appropriate to write	+	Wir bringen zuerst
-	<math>1+i\sqrt{3}</math> ~~and~~ <math>\text{1}+i</math> in ~~polar form right from the beginning and to carry out all calculations in polar form~~.	+	<math>1+i\sqrt{3}</math> und <math>\text{1}+i</math> auf Polarform, und rechnen weiterhin in Polarform.

	<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>		<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>

-	~~We obtain~~	+	Wir erhalten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 11:		Zeile 11:
	\end{align}</math>}}		\end{align}</math>}}

-	~~and~~	+	und also

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 20:		Zeile 20:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Finally, de Moivre's formula gives~~	+	Durch den Moivreschen Satz erhalten wir schließlich

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

Tekbot: Solution 3.3:1d moved to Lösung 3.3:1d: Robot: moved page

2009-03-11T10:43:35Z

Solution 3.3:1d moved to Lösung 3.3:1d: Robot: moved page

← Nächstältere Version

Version vom 10:43, 11. Mär. 2009

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

2009-03-10T13:11:14Z

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

← Nächstältere Version		Version vom 13:11, 10. Mär. 2009
Zeile 6:		Zeile 6:
	We obtain		We obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]		1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]
	1+i &= \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)		1+i &= \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)
Zeile 13:		Zeile 13:
	and		and

-	{{~~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]
Zeile 22:		Zeile 22:
	Finally, de Moivre's formula gives		Finally, de Moivre's formula gives

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12}		\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12}
	&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt]		&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt]

Tek um 09:26, 30. Okt. 2008

2008-10-30T09:26:26Z

← Nächstältere Version		Version vom 09:26, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	Because we are going to raise something to the power	+	Because we are going to raise something to the power 12, the base in the expression should be written in polar form. In turn, the base consists of a quotient which is advantageous to calculate in polar form. Thus, it seems appropriate to write
-	~~<math>\text{~~12~~}</math>~~, the base in the expression should be written in polar form. In turn, the base consists of a quotient which it is advantageous to calculate in polar form. Thus, it seems appropriate to write	+	<math>1+i\sqrt{3}</math> and <math>\text{1}+i</math> in polar form right from the beginning and to carry out all calculations in polar form.
-	<math>1+i\sqrt{3}</math>	+
-	and	+
-	<math>\text{1}+i</math> in polar form right from the beginning and to carry out all calculations in polar form.	+
-		+
-		+
-	~~[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]~~	+

		+	<center>[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]</center>

	We obtain		We obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\\[5pt]
-	& 1+i\sqrt{3}=2\~~left~~( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \~~right~~) \\	+	1+i &= \sqrt{2}\Bigl(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\Bigr)
-	~~& \text{~~1}+i=\sqrt{2}\~~left~~( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \~~right~~) \\	+	\end{align}</math>}}
-	\end{align}</math>	+
-		+

	and		and

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{1+i\sqrt{3}}{1+i}
-	& \frac{1+i\sqrt{3}}~~{\text~~{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>	+
-		+

	Finally, de Moivre's formula gives		Finally, de Moivre's formula gives

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\Bigl(\frac{1+i\sqrt{3}}{1+i}\Bigr)^{12}
-	& \~~left~~( \frac{1+i\sqrt{3}}~~{\text~~{1}+i} \~~right~~)^{12}=\~~left~~( \sqrt{2} \~~right~~)^{12}\~~left~~( \cos 12\~~centerdot~~ \frac{\pi }{12}+i\sin 12\~~centerdot~~ \frac{\pi }{12} \~~right~~) \\	+	&= \bigl(\sqrt{2}\,\bigr)^{12}\Bigl(\cos\Bigl(12\cdot\frac{\pi}{12}\Bigr) + i\sin\Bigl(12\cdot\frac{\pi}{12}\Bigr)\Bigr)\\[5pt]
-	& =2^{~~\frac{~~1}{2}\~~centerdot~~ 12}~~\left~~( \cos \pi +i\sin \pi ~~\right~~) \\	+	&= 2^{(1/2)\cdot 12}(\cos\pi + i\sin\pi)\\[5pt]
-	& =2^{6}\~~centerdot \left~~( -1+i\~~centerdot~~ 0 ~~\right~~) \\	+	&= 2^6\cdot (-1+i\cdot 0)\\[5pt]
-	& =-64 \\	+	&= -64\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian um 07:49, 24. Okt. 2008

2008-10-24T07:49:42Z

← Nächstältere Version		Version vom 07:49, 24. Okt. 2008
Zeile 1:		Zeile 1:
-	{~~{NAVCONTENT_START}~~}	+	Because we are going to raise something to the power
-	<~~center~~> ~~[[Image:3_3_1d-~~1~~(2).gif]]~~ </~~center~~>	+	<math>\text{12}</math>, the base in the expression should be written in polar form. In turn, the base consists of a quotient which it is advantageous to calculate in polar form. Thus, it seems appropriate to write
-	~~{{NAVCONTENT_STOP}}~~	+	<math>1+i\sqrt{3}</math>
-	{~~{NAVCONTENT_START~~}}	+	and
-	<~~center~~> ~~[[Image:3_3_1d-2(2)~~.~~gif]] </center>~~	+	<math>\text{1}+i</math> in polar form right from the beginning and to carry out all calculations in polar form.
-	~~{{NAVCONTENT_STOP}}~~	+

	[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]		[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]
		+
		+
		+	We obtain
		+
		+
		+	<math>\begin{align}
		+	& 1+i\sqrt{3}=2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\
		+	& \text{1}+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \\
		+	\end{align}</math>
		+
		+
		+	and
		+
		+
		+	<math>\begin{align}
		+	& \frac{1+i\sqrt{3}}{\text{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>
		+
		+
		+	Finally, de Moivre's formula gives
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{1+i\sqrt{3}}{\text{1}+i} \right)^{12}=\left( \sqrt{2} \right)^{12}\left( \cos 12\centerdot \frac{\pi }{12}+i\sin 12\centerdot \frac{\pi }{12} \right) \\
		+	& =2^{\frac{1}{2}\centerdot 12}\left( \cos \pi +i\sin \pi \right) \\
		+	& =2^{6}\centerdot \left( -1+i\centerdot 0 \right) \\
		+	& =-64 \\
		+	\end{align}</math>

Tekbot: Lösning 3.3:1d moved to Solution 3.3:1d: Robot: moved page

2008-09-16T15:04:01Z

Lösning 3.3:1d moved to Solution 3.3:1d: Robot: moved page

← Nächstältere Version

Version vom 15:04, 16. Sep. 2008

Madde um 12:47, 21. Aug. 2008

2008-08-21T12:47:34Z

← Nächstältere Version		Version vom 12:47, 21. Aug. 2008
Zeile 5:		Zeile 5:
	<center> [[Image:3_3_1d-2(2).gif]] </center>		<center> [[Image:3_3_1d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	[[Image:3_3_1_d.gif]] [[Image:3_3_1_d_text.gif]]

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

2008-08-21T08:11:44Z

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

← Nächstältere Version		Version vom 08:11, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_1d-1(2).gif]] </center>	+	<center> [[Image:3_3_1d-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_1d-2(2).gif]] </center>	+	<center> [[Image:3_3_1d-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}