Lösung 3.2:6c - Versionsgeschichte

Moni um 16:39, 22. Aug. 2009

2009-08-22T16:39:44Z

← Nächstältere Version		Version vom 16:39, 22. Aug. 2009
Zeile 1:		Zeile 1:
	Wir brauchen nur den Betrag und das Argument um die Zahl auf Polarform zu bringen.		Wir brauchen nur den Betrag und das Argument um die Zahl auf Polarform zu bringen.

-	Den ~~betrag~~ erhalten wir durch die Formel	+	Den Betrag erhalten wir durch die Formel

	{{Abgesetzte Formel\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}

-	Weiterhin ~~lieht~~ die Zahl in dritten Quadrant, und definieren wir ein Dreieck wie in der Figur, können wir zuerst den Winkel <math>\alpha</math> bestimmen. Addieren wir <math>\pi</math> zu ~~diesen~~ Winkel, erhalten wir das Argument.	+	Weiterhin liegt die Zahl in dritten Quadrant, und definieren wir ein Dreieck wie in der Figur, können wir zuerst den Winkel <math>\alpha</math> bestimmen. Addieren wir <math>\pi</math> zu diesem Winkel, erhalten wir das Argument.

	<center>[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]</center>		<center>[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]</center>

Martin um 13:09, 13. Mai 2009

2009-05-13T13:09:39Z

← Nächstältere Version		Version vom 13:09, 13. Mai 2009
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-	~~We need to know the number's magnitude and argument, in order to be able to write down its polar form~~.	+	Wir brauchen nur den Betrag und das Argument um die Zahl auf Polarform zu bringen.

-	~~We obtain the number's magnitude as its distance to the origin; it is obtained by the distance formula,~~	+	Den betrag erhalten wir durch die Formel

	{{Abgesetzte Formel\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}

-	~~Furthermore~~, ~~the number lies~~ in ~~the third quadrant (negative real and imaginary parts) and if we introduce an auxiliary triangle in that quadrant~~, ~~we can determine the angle that the number makes with the negative real axis~~. ~~Then, we get the number's argument by adding the angle~~ <math>\pi</math>.	+	Weiterhin lieht die Zahl in dritten Quadrant, und definieren wir ein Dreieck wie in der Figur, können wir zuerst den Winkel <math>\alpha</math> bestimmen. Addieren wir <math>\pi</math> zu diesen Winkel, erhalten wir das Argument.

	<center>[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]</center>		<center>[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]</center>

-	~~The number's polar form is thus~~	+	Die Polarform der Zahl ist daher

	{{Abgesetzte Formel\|\|<math>4\sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \Bigr)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>4\sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \Bigr)\,\textrm{.}</math>}}

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

2009-03-11T10:41:15Z

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

← Nächstältere Version

Version vom 10:41, 11. Mär. 2009

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

2009-03-10T13:10:04Z

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

← Nächstältere Version		Version vom 13:10, 10. Mär. 2009
Zeile 3:		Zeile 3:
	We obtain the number's magnitude as its distance to the origin; it is obtained by the distance formula,		We obtain the number's magnitude as its distance to the origin; it is obtained by the distance formula,

-	{{~~Displayed math~~\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}

	Furthermore, the number lies in the third quadrant (negative real and imaginary parts) and if we introduce an auxiliary triangle in that quadrant, we can determine the angle that the number makes with the negative real axis. Then, we get the number's argument by adding the angle <math>\pi</math>.		Furthermore, the number lies in the third quadrant (negative real and imaginary parts) and if we introduce an auxiliary triangle in that quadrant, we can determine the angle that the number makes with the negative real axis. Then, we get the number's argument by adding the angle <math>\pi</math>.
Zeile 11:		Zeile 11:
	The number's polar form is thus		The number's polar form is thus

-	{{~~Displayed math~~\|\|<math>4\sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \Bigr)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>4\sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \Bigr)\,\textrm{.}</math>}}

Tek um 13:18, 29. Okt. 2008

2008-10-29T13:18:51Z

← Nächstältere Version		Version vom 13:18, 29. Okt. 2008
Zeile 1:		Zeile 1:
	We need to know the number's magnitude and argument, in order to be able to write down its polar form.		We need to know the number's magnitude and argument, in order to be able to write down its polar form.

-	We obtain the number's magnitude as its distance to the origin; it is obtained by the distance formula:	+	We obtain the number's magnitude as its distance to the origin; it is obtained by the distance formula,

		+	{{Displayed math\|\|<math>\|-4-4i\| = \sqrt{(-4)^2+(-4)^2} = \sqrt{16+16} = \sqrt{16\cdot 2} = 4\sqrt{2}\,\textrm{.}</math>}}

-	<math>\~~left\| -4-4i \right\|=\sqrt{\left( -4 \right)^{2}+\left( -4 \right)^{2}}=\sqrt{16+16}=\sqrt{16\centerdot 2}=4\sqrt{2}~~</math>	+	Furthermore, the number lies in the third quadrant (negative real and imaginary parts) and if we introduce an auxiliary triangle in that quadrant, we can determine the angle that the number makes with the negative real axis. Then, we get the number's argument by adding the angle <math>\pi</math>.

-	Furthermore, the number lies in the third quadrant (negative real and imaginary parts) and if we introduce an auxiliary triangle in that quadrant, we can determine the angle that the number makes with the negative real axis. Then, we get the number's argument by adding the angle	+	<center>[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]</center>
-	<~~math~~>~~\pi </math>.~~	+
-		+
-		+
-	[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]	+
-	<~~!-- texten är i en bild och figuern i en... --~~>	+

	The number's polar form is thus		The number's polar form is thus

-		+	{{Displayed math\|\|<math>4\sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4} \Bigr)\,\textrm{.}</math>}}
-	<math>4\sqrt{2}\~~left~~( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \~~right~~)</math>	+

Ian um 10:22, 23. Okt. 2008

2008-10-23T10:22:58Z

← Nächstältere Version		Version vom 10:22, 23. Okt. 2008
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-	{{~~NAVCONTENT_START~~}}	+	We need to know the number's magnitude and argument, in order to be able to write down its polar form.
-	<~~center~~> ~~[[Image:3_2_6c~~.~~gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+	We obtain the number's magnitude as its distance to the origin; it is obtained by the distance formula:
		+
		+
		+	<math>\left\| -4-4i \right\|=\sqrt{\left( -4 \right)^{2}+\left( -4 \right)^{2}}=\sqrt{16+16}=\sqrt{16\centerdot 2}=4\sqrt{2}</math>
		+
		+	Furthermore, the number lies in the third quadrant (negative real and imaginary parts) and if we introduce an auxiliary triangle in that quadrant, we can determine the angle that the number makes with the negative real axis. Then, we get the number's argument by adding the angle
		+	<math>\pi </math>.
		+

	[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]		[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]
	<!-- texten är i en bild och figuern i en... -->		<!-- texten är i en bild och figuern i en... -->
		+
		+	The number's polar form is thus
		+
		+
		+	<math>4\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right)</math>

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

2008-09-16T15:01:41Z

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

← Nächstältere Version

Version vom 15:01, 16. Sep. 2008

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

2008-08-21T08:10:34Z

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

← Nächstältere Version		Version vom 08:10, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_6c.gif]] </center>	+	<center> [[Image:3_2_6c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[~~Bild~~:3_2_6_c1.gif]] [[~~Bild~~:3_2_6_c2.gif]]	+	[[Image:3_2_6_c1.gif]] [[Image:3_2_6_c2.gif]]
	<!-- texten är i en bild och figuern i en... -->		<!-- texten är i en bild och figuern i en... -->

Madde um 16:10, 19. Aug. 2008

2008-08-19T16:10:50Z

← Nächstältere Version		Version vom 16:10, 19. Aug. 2008
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[Bild:3_2_6_c1.gif]] [[Bild:3_2_6_c2.gif]]	+	[[Bild:3_2_6_c1.gif]] [[Bild:3_2_6_c2.gif]]
		+	<!-- texten är i en bild och figuern i en... -->

Madde um 16:09, 19. Aug. 2008

2008-08-19T16:09:10Z

← Nächstältere Version		Version vom 16:09, 19. Aug. 2008
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	<center> [[Bild:3_2_6c.gif]] </center>		<center> [[Bild:3_2_6c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	[[Bild:3_2_6_c1.gif]] [[Bild:3_2_6_c2.gif]]