Lösung 3.1:4b - Versionsgeschichte

Moni um 12:15, 22. Aug. 2009

2009-08-22T12:15:07Z

← Nächstältere Version		Version vom 12:15, 22. Aug. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}

-	Also müssen wir den Bruch auf der linken Seite berechnen. Wir erweitern den Bruch mit ~~den~~ konjugiert komplexen Nenner,	+	Also müssen wir den Bruch auf der linken Seite berechnen. Wir erweitern den Bruch mit dem konjugiert komplexen Nenner,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	Wir substituieren <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> in der Ursprünglichen Gleichung, um zu kontrollieren ~~dass~~ wir richtig gerechnet haben,	+	Wir substituieren <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> in der Ursprünglichen Gleichung, um zu kontrollieren, ob wir richtig gerechnet haben,

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

Martin um 10:25, 12. Mai 2009

2009-05-12T10:25:42Z

← Nächstältere Version		Version vom 10:25, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~If we divide both sides by~~ <math>2-i</math>, ~~we obtain~~ <math>z</math> ~~by itself on the left-hand side~~,	+	Wenn wir beide Seiten durch <math>2-i</math> dividieren, erhalten wir <math>z</math> auf der linken Seite,

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

-	~~It remains to calculate the quotient on the right-hand side~~. ~~We multiply top and bottom by the complex conjugate of the denominator~~,	+	Also müssen wir den Bruch auf der linken Seite berechnen. Wir erweitern den Bruch mit den konjugiert komplexen Nenner,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Also, we substitute~~ <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> ~~into the original equation to assure ourselves that we have calculated correctly~~,	+	Wir substituieren <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> in der Ursprünglichen Gleichung, um zu kontrollieren dass wir richtig gerechnet haben,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~LHS~~}	+	\text{Linke Seite}
	&= (2-i)z\\[5pt]		&= (2-i)z\\[5pt]
	&= (2-i)\Bigl(\frac{4}{5}+\frac{7}{5}\,i\bigr)\\[5pt]		&= (2-i)\Bigl(\frac{4}{5}+\frac{7}{5}\,i\bigr)\\[5pt]
Zeile 24:		Zeile 24:
	&= \frac{15}{5} + \frac{10}{5}\,i\\[5pt]		&= \frac{15}{5} + \frac{10}{5}\,i\\[5pt]
	&= 3+2i\\[5pt]		&= 3+2i\\[5pt]
-	&= \text{~~RHS~~.}\end{align}</math>}}	+	&= \text{Rechte Seite.}\end{align}</math>}}

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

2009-03-11T10:32:35Z

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

← Nächstältere Version

Version vom 10:32, 11. Mär. 2009

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

2009-03-10T13:06:44Z

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

← Nächstältere Version		Version vom 13:06, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,		If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,

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

	It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,		It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]		z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
	&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]		&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
Zeile 15:		Zeile 15:
	Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,		Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{LHS}		\text{LHS}
	&= (2-i)z\\[5pt]		&= (2-i)z\\[5pt]

Tek um 07:27, 30. Okt. 2008

2008-10-30T07:27:26Z

← Nächstältere Version		Version vom 07:27, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,
-	If we divide both sides by 2-i, we obtain z by itself on the left-hand side:	+

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

-	~~<math>z=\frac{3+2i}{2~~-i}.~~</math>~~	+	It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,

		+	{{Displayed math\|\|<math>\begin{align}
		+	z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
		+	&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
		+	&= \frac{6+3i+4i-2}{4+1}\\[5pt]
		+	&= \frac{4+7i}{5}\\[5pt]
		+	&= \frac{4}{5}+\frac{7}{5}\,i\,\textrm{.}
		+	\end{align}</math>}}

-	~~It remains~~ to ~~calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the numerator:~~	+	Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}~~z&=~~ \~~frac~~{~~(3+2i)(2+i)~~}{(2-i)~~(2+i)}=\frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\~~	+	\text{LHS}
-	~~&=\frac{6+3i+4i-2}{4+1}=\frac{4+7i}{5}=\frac{4}{5}+\frac{7}{5}i\end{align}</math>~~	+	&= (2-i)z\\[5pt]
-		+	&= (2-i)\Bigl(\frac{4}{5}+\frac{7}{5}\,i\bigr)\\[5pt]
-		+	&= 2\cdot\frac{4}{5} + 2\cdot\frac{7}{5}\,i - i\cdot\frac{4}{5} - i\cdot\frac{7}{5}\,i\\[5pt]
-	~~Also, we substitute <math>~~z=\~~frac{4}{5}+~~\~~frac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly:~~	+	&= \frac{8}{5} + \frac{14}{5}\,i - \frac{4}{5}\,i + \frac{7}{5}\\[5pt]
-		+	&= \frac{8+7}{5} + \frac{14-4}{5}\,i\\[5pt]
-		+	&= \frac{15}{5} + \frac{10}{5}\,i\\[5pt]
-	~~<math>\begin{align}LHS~~ &= (2-i)~~z=(2-i)~~(\frac{4}{5}+\frac{7}{5}i)\\	+	&= 3+2i\\[5pt]
-	&=2\cdot\frac{4}{5}-i\cdot\frac{4}{5}+2\cdot \frac{7}{5}i -i\cdot \frac{7}{5}i\\	+	&= \text{RHS.}\end{align}</math>}}
-	&=\frac{8}{5}-\frac{4}{5}i+\frac{14}{5}i+\frac{7}{5} = \frac{8+7}{5}+\frac{14-4}{5}i\\	+
-	&=\frac{15}{5}+\frac{10}{5}i=3+2i=RHS.\end{align}</math>	+
-		+
-	~~{{NAVCONTENT_STOP~~}}	+

Jamesjmnewton um 10:53, 23. Sep. 2008

2008-09-23T10:53:33Z

← Nächstältere Version		Version vom 10:53, 23. Sep. 2008
Zeile 13:		Zeile 13:


-	Also, we substitute into the original equation to assure ourselves that we have calculated correctly:	+	Also, we substitute <math>z=\frac{4}{5}+\frac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly:

Jamesjmnewton um 10:44, 23. Sep. 2008

2008-09-23T10:44:39Z

← Nächstältere Version		Version vom 10:44, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image~~:~~3_1_4b~~.~~gif]]~~ </~~center~~>	+	If we divide both sides by 2-i, we obtain z by itself on the left-hand side:
		+
		+
		+	<math>z=\frac{3+2i}{2-i}.</math>
		+
		+
		+	It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the numerator:
		+
		+
		+	<math>\begin{align}z&= \frac{(3+2i)(2+i)}{(2-i)(2+i)}=\frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\
		+	&=\frac{6+3i+4i-2}{4+1}=\frac{4+7i}{5}=\frac{4}{5}+\frac{7}{5}i\end{align}</math>
		+
		+
		+	Also, we substitute into the original equation to assure ourselves that we have calculated correctly:
		+
		+
		+	<math>\begin{align}LHS &= (2-i)z=(2-i)(\frac{4}{5}+\frac{7}{5}i)\\
		+	&=2\cdot\frac{4}{5}-i\cdot\frac{4}{5}+2\cdot \frac{7}{5}i -i\cdot \frac{7}{5}i\\
		+	&=\frac{8}{5}-\frac{4}{5}i+\frac{14}{5}i+\frac{7}{5} = \frac{8+7}{5}+\frac{14-4}{5}i\\
		+	&=\frac{15}{5}+\frac{10}{5}i=3+2i=RHS.\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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

2008-09-16T14:53:01Z

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

← Nächstältere Version

Version vom 14:53, 16. Sep. 2008

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

2008-08-21T08:06:14Z

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

← Nächstältere Version		Version vom 08:06, 21. Aug. 2008
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-	<center> [[~~Bild~~:3_1_4b.gif]] </center>	+	<center> [[Image:3_1_4b.gif]] </center>
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_4b.gif {{NAVCONTENT_STOP}}

2008-04-07T10:52:35Z

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