Lösung 3.1:4e - Versionsgeschichte

Moni um 12:18, 22. Aug. 2009

2009-08-22T12:18:12Z

← Nächstältere Version		Version vom 12:18, 22. Aug. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}

-	Dies bedeutet aber dass wir eine neue Gleichung haben, die nicht äquivalent mit der ursprünglichen Gleichung ist. Wenn die neue Gleichung eine Lösung <math>z=-i</math> hat, kann dies unmöglich eine Lösung der ursprünglichen Gleichung sein, nachdem der Nenner dann 0 wäre.	+	Dies bedeutet aber, dass wir eine neue Gleichung haben, die nicht äquivalent mit der ursprünglichen Gleichung ist. Wenn die neue Gleichung eine Lösung <math>z=-i</math> hat, kann dies unmöglich eine Lösung der ursprünglichen Gleichung sein, nachdem der Nenner dann 0 wäre.

	Wir erweitern die rechte Seite der neuen Gleichung,		Wir erweitern die rechte Seite der neuen Gleichung,

Martin um 10:38, 12. Mai 2009

2009-05-12T10:38:37Z

← Nächstältere Version		Version vom 10:38, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~If we multiply both sides by~~ <math>z+i</math>, ~~we avoid having~~ <math>z</math> ~~in the denominator~~,	+	Wir multiplizieren beide Seiten mit <math>z+i</math>, und werden dadurch <math>z</math> im Nenner los,

	{{Abgesetzte Formel\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}

-	~~At the same time~~, ~~this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation~~. ~~Should the new equation show itself to have~~ <math>z=-i</math> ~~as a solution~~, ~~then we must ignore that solution~~, ~~because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero)~~.	+	Dies bedeutet aber dass wir eine neue Gleichung haben, die nicht äquivalent mit der ursprünglichen Gleichung ist. Wenn die neue Gleichung eine Lösung <math>z=-i</math> hat, kann dies unmöglich eine Lösung der ursprünglichen Gleichung sein, nachdem der Nenner dann 0 wäre.

-	~~We expand the right-hand side in the new equation~~,	+	Wir erweitern die rechte Seite der neuen Gleichung,

	{{Abgesetzte Formel\|\|<math>iz+1 = 3z+3i+iz-1\,,</math>}}		{{Abgesetzte Formel\|\|<math>iz+1 = 3z+3i+iz-1\,,</math>}}

-	~~and move all the terms in~~ <math>z</math> ~~over to the left~~-~~hand side and the constants to the right-hand side~~,	+	und ziehen alle <math>z</math>-Terme zur linken Seite, und alle Konstanten zur rechten Seite,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 16:		Zeile 16:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Then~~, ~~we obtain~~	+	Wir erhalten also,

	{{Abgesetzte Formel\|\|<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}

-	~~It is a little troublesome to divide two complex numbers~~, ~~so we will therefore not check whether <math>z=\tfrac{2}{3}-i</math> is a solution to the original equation~~, ~~but satisfy ourselves with substituting into the equation~~ <math>iz+1=(3+i)(z+i)</math>,	+	Nachdem es recht mühsam ist, kompleze Zahlen mit einander zu dividieren, kontrollieren wir nicht die Lösung in der ursprünglichen Gleichung, sondern in der Gleichung <math>iz+1=(3+i)(z+i)</math>,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~LHS~~} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]	+	\text{Linke Seite} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]
-	\text{~~RHS~~} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}	+	\text{Rechte Seite} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tekbot: Solution 3.1:4e moved to Lösung 3.1:4e: Robot: moved page

2009-03-11T10:33:35Z

Solution 3.1:4e moved to Lösung 3.1:4e: Robot: moved page

← Nächstältere Version

Version vom 10:33, 11. Mär. 2009

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

2009-03-10T13:07:14Z

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

← Nächstältere Version		Version vom 13:07, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we multiply both sides by <math>z+i</math>, we avoid having <math>z</math> in the denominator,		If we multiply both sides by <math>z+i</math>, we avoid having <math>z</math> in the denominator,

-	{{~~Displayed math~~\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}

	At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have <math>z=-i</math> as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).		At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have <math>z=-i</math> as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).
Zeile 7:		Zeile 7:
	We expand the right-hand side in the new equation,		We expand the right-hand side in the new equation,

-	{{~~Displayed math~~\|\|<math>iz+1 = 3z+3i+iz-1\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>iz+1 = 3z+3i+iz-1\,,</math>}}

	and move all the terms in <math>z</math> over to the left-hand side and the constants to the right-hand side,		and move all the terms in <math>z</math> over to the left-hand side and the constants to the right-hand side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	iz-3z-iz &= 3i-1-1\,,\\[5pt]		iz-3z-iz &= 3i-1-1\,,\\[5pt]
	-3z &= -2+3i\,\textrm{.}		-3z &= -2+3i\,\textrm{.}
Zeile 18:		Zeile 18:
	Then, we obtain		Then, we obtain

-	{{~~Displayed math~~\|\|<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}

	It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\tfrac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>,		It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\tfrac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{LHS} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]		\text{LHS} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]
	\text{RHS} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}		\text{RHS} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 08:20, 30. Okt. 2008

2008-10-30T08:20:22Z

← Nächstältere Version		Version vom 08:20, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	If we multiply both sides by <math>z+i</math>, we avoid having <math>z</math> in the denominator,

-	~~If we multiply both sides by~~ z+i, ~~we avoid having z in the denominator:~~	+	{{Displayed math\|\|<math>iz+1=(3+i)(z+i)\,\textrm{.}</math>}}

-		+	At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have <math>z=-i</math> as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).
-	~~<math>iz+1=(3+i)(z+i).</math>~~	+
-		+
-		+
-	At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have z=-i as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).	+

	We expand the right-hand side in the new equation,		We expand the right-hand side in the new equation,

		+	{{Displayed math\|\|<math>iz+1 = 3z+3i+iz-1\,,</math>}}

-	<math>~~iz+1=3z+3i+iz-1~~</math>	+	and move all the terms in <math>z</math> over to the left-hand side and the constants to the right-hand side,
-		+
-		+
-	~~and move all the terms in z~~ over to the left-hand side and the constants to the right-hand side:	+
-		+
-		+
-	~~<math>\begin{align}iz-3z-iz &= 3i-1-1~~,\\	+
-	~~-3z &= -2+3i.\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	iz-3z-iz &= 3i-1-1\,,\\[5pt]
		+	-3z &= -2+3i\,\textrm{.}
		+	\end{align}</math>}}

	Then, we obtain		Then, we obtain

		+	{{Displayed math\|\|<math>z = \frac{-2+3i}{-3} = \frac{2}{3}-i\,\textrm{.}</math>}}

-	~~<math>z=\frac{-2+3i}{-3}=\frac{2}{3}-i.</math>~~	+	It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\tfrac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>,
-		+
-	It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\~~frac~~{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>:	+
-		+
-		+
-	~~<math>\begin{align}LHS &=iz+1=i(\frac{2}{3}-i)+1=\frac{2}{3}\cdot i+1+1=2+\frac{2}{3}i~~,\\	+
-	~~RHS &= (3+i)(z+i)=(3+i)(\frac{2}{3}-i+i)=(3+i)\frac{2}{3}=2+\frac{2}{3}i.\end{align}</math>~~	+

-	{{~~NAVCONTENT_STOP~~}}	+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS} &= iz+1 = i(\tfrac{2}{3}-i)+1 = \tfrac{2}{3}\cdot i+1+1 = 2+\tfrac{2}{3}i,\\[5pt]
		+	\text{RHS} &= (3+i)(z+i) = (3+i)(\tfrac{2}{3}-i+i) = (3+i)\tfrac{2}{3} = 2+\tfrac{2}{3}i\,\textrm{.}
		+	\end{align}</math>}}

Jamesjmnewton um 11:35, 23. Sep. 2008

2008-09-23T11:35:05Z

← Nächstältere Version		Version vom 11:35, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_4e-~~1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP}~~}	+	If we multiply both sides by z+i, we avoid having z in the denominator:
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image~~:~~3_1_4e~~-2(2).~~gif]]~~ </~~center~~>	+
		+	<math>iz+1=(3+i)(z+i).</math>
		+
		+
		+	At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have z=-i as a solution, then we must ignore that solution, because our initial equation cannot possibly have <math>z=-i</math> as a solution (the denominator of the left-hand side becomes zero).
		+
		+	We expand the right-hand side in the new equation,
		+
		+
		+	<math>iz+1=3z+3i+iz-1</math>
		+
		+
		+	and move all the terms in z over to the left-hand side and the constants to the right-hand side:
		+
		+
		+	<math>\begin{align}iz-3z-iz &= 3i-1-1,\\
		+	-3z &= -2+3i.\end{align}</math>
		+
		+
		+	Then, we obtain
		+
		+
		+	<math>z=\frac{-2+3i}{-3}=\frac{2}{3}-i.</math>
		+
		+	It is a little troublesome to divide two complex numbers, so we will therefore not check whether <math>z=\frac{2}{3}-i</math> is a solution to the original equation, but satisfy ourselves with substituting into the equation <math>iz+1=(3+i)(z+i)</math>:
		+
		+
		+	<math>\begin{align}LHS &=iz+1=i(\frac{2}{3}-i)+1=\frac{2}{3}\cdot i+1+1=2+\frac{2}{3}i,\\
		+	RHS &= (3+i)(z+i)=(3+i)(\frac{2}{3}-i+i)=(3+i)\frac{2}{3}=2+\frac{2}{3}i.\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot: Lösning 3.1:4e moved to Solution 3.1:4e: Robot: moved page

2008-09-16T14:54:01Z

Lösning 3.1:4e moved to Solution 3.1:4e: Robot: moved page

← Nächstältere Version

Version vom 14:54, 16. Sep. 2008

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

2008-08-21T08:06:44Z

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

← Nächstältere Version		Version vom 08:06, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_4e-1(2).gif]] </center>	+	<center> [[Image:3_1_4e-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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-	<center> [[~~Bild~~:3_1_4e-2(2).gif]] </center>	+	<center> [[Image:3_1_4e-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_4e-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_1_4e-2(2).gif {{NAVCONTENT_STOP}}

2008-04-07T10:54:38Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_4e-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_1_4e-2(2).gif </center> {{NAVCONTENT_STOP}}

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