Lösung 3.1:4d - Versionsgeschichte

Martin um 10:32, 12. Mai 2009

Martin — Tue, 12 May 2009 10:32:30 GMT

← Nächstältere Version		Version vom 10:32, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~In the equation,~~ <math>z</math> ~~occurs only as~~ <math>\bar{z}</math> ~~and~~, ~~to begin with, we can therefore treat~~ <math>\bar{z}</math> ~~as unknown.~~	+	Nachdem <math>z</math> nur in Termen von <math>\bar{z}</math> vorkommt, können wir zuerst <math>\bar{z}</math> als Unbekannte behandeln,

-	~~Divide both sides by~~ <math>2+i</math>,	+	Wir dividieren beide Seiten durch <math>2+i</math>,

	{{Abgesetzte Formel\|\|<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}		{{Abgesetzte Formel\|\|<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}

-	~~and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator~~,	+	und berechnen die rechte Seite indem wir den Bruch mit den konjugiert komplexen Nenner erweitern,

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

-	~~This means that~~ <math>z=\tfrac{3}{5}-\tfrac{1}{5}i\,</math>.	+	Also ist <math>z=\tfrac{3}{5}-\tfrac{1}{5}i\,</math>.

-	~~We check that~~ <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> ~~satisfies the original equation~~,	+	Wir kontrollieren wie immer dass <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> die ursprüngliche Gleichung erfüllt,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~LHS~~}	+	\text{Linke Seite}
	&= (2+i)\bar{z}		&= (2+i)\bar{z}
	= (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)}		= (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)}
Zeile 29:		Zeile 29:
	&=\frac{6-1}{5}+\frac{2+3}{5}\,i		&=\frac{6-1}{5}+\frac{2+3}{5}\,i
	= 1+i		= 1+i
-	= \text{~~RHS~~}\,\textrm{.}	+	= \text{Rechte Seite}\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

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

Tekbot — Wed, 11 Mar 2009 10:33:15 GMT

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

← Nächstältere Version

Version vom 10:33, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:07:04 GMT

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

← Nächstältere Version		Version vom 13:07, 10. Mär. 2009
Zeile 3:		Zeile 3:
	Divide both sides by <math>2+i</math>,		Divide both sides by <math>2+i</math>,

-	{{~~Displayed math~~\|\|<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}

	and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,		and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\bar{z}		\bar{z}
	&= \frac{(1+i)(2-i)}{(2+i)(2-i)}		&= \frac{(1+i)(2-i)}{(2+i)(2-i)}
Zeile 20:		Zeile 20:
	We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,		We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,

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

Tek um 08:09, 30. Okt. 2008

Tek — Thu, 30 Oct 2008 08:09:06 GMT

← Nächstältere Version		Version vom 08:09, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}
	In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.		In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.

	Divide both sides by <math>2+i</math>,		Divide both sides by <math>2+i</math>,

		+	{{Displayed math\|\|<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}

-	~~<math>\bar{z}=\frac{1+i}{2+i}</math>~~	+	and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,

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

-	~~and calculate the quotient on the right~~-~~hand side by multiplying top and bottom by the complex conjugate of the numerator:~~	+	This means that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i\,</math>.

		+	We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,

-	~~<math>\begin~~{~~align}\bar~~{~~z}&=\frac{(1+i)(2-i)}{(2+i)(2-i)}=\frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\~~	+	{{Displayed math\|\|<math>\begin{align}
-	~~&=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}</~~math>	+	\text{LHS}
-		+	&= (2+i)\bar{z}
-		+	= (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)}
-	~~This means that~~ <math>z=\~~frac~~{3}~~{5}-\frac{1}{5}i.</math>~~	+	= (2+i)\Bigl(\frac{3}{5}+\frac{1}{5}\,i\Bigr)\\[5pt]
-		+	&= 2\cdot\frac{3}{5}+2\cdot\frac{1}{5}\,i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}\,i
-	~~We check that <math>z=~~\~~frac~~{3}~~{5}-\frac{1}{5}i</math> satisfies the original equation:~~	+	= \frac{6}{5}+\frac{2}{5}\,i+\frac{3}{5}\,i-\frac{1}{5}\\[5pt]
-		+	&=\frac{6-1}{5}+\frac{2+3}{5}\,i
-		+	= 1+i
-	~~<math>\begin{align}LHS~~ &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\	+	= \text{RHS}\,\textrm{.}
-	&=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\	+	\end{align}</math>}}
-	&=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}</math>	+
-	~~{{NAVCONTENT_STOP~~}}	+

Jamesjmnewton um 11:03, 23. Sep. 2008

Jamesjmnewton — Tue, 23 Sep 2008 11:03:33 GMT

← Nächstältere Version		Version vom 11:03, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	In the equation, z occurs only as and, to begin with, we can therefore ~~treatas~~ unknown.	+	In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.

-	Divide both sides by 2+i,	+	Divide both sides by <math>2+i</math>,
		+
		+
		+	<math>\bar{z}=\frac{1+i}{2+i}</math>

-	EQ1

	and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:		and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:

-	EQ2

-	This means that .	+	<math>\begin{align}\bar{z}&=\frac{(1+i)(2-i)}{(2+i)(2-i)}=\frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\
		+	&=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}</math>
		+
		+
		+	This means that <math>z=\frac{3}{5}-\frac{1}{5}i.</math>

-	We check that satisfies the original equation:	+	We check that <math>z=\frac{3}{5}-\frac{1}{5}i</math> satisfies the original equation:

-	EQ3

		+	<math>\begin{align}LHS &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\
		+	&=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\
		+	&=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}</math>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Jamesjmnewton um 10:52, 23. Sep. 2008

Jamesjmnewton — Tue, 23 Sep 2008 10:52:49 GMT

← Nächstältere Version		Version vom 10:52, 23. Sep. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	~~<center> [[Image~~:~~3_1_4d~~.~~gif]] </center>~~	+	In the equation, z occurs only as and, to begin with, we can therefore treatas unknown.
		+
		+	Divide both sides by 2+i,
		+
		+	EQ1
		+
		+	and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:
		+
		+	EQ2
		+
		+	This means that .
		+
		+	We check that satisfies the original equation:
		+
		+	EQ3
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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

Tekbot — Tue, 16 Sep 2008 14:53:41 GMT

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

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Version vom 14:53, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:06:34 GMT

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

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-	<center> [[~~Bild~~:3_1_4d.gif]] </center>	+	<center> [[Image:3_1_4d.gif]] </center>
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_4d.gif {{NAVCONTENT_STOP}}

Oskar — Mon, 07 Apr 2008 10:53:57 GMT

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Neue Seite

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