Lösung 3.1:4f - Versionsgeschichte

Moni um 12:23, 22. Aug. 2009

2009-08-22T12:23:44Z

← Nächstältere Version		Version vom 12:23, 22. Aug. 2009
Zeile 1:		Zeile 1:
-	Nachdem <math>z</math> und <math>\bar{z}</math> in der Gleichung, ~~we cannot use~~ <math>z</math> (oder <math>\bar{z}</math>) lösen, ~~und wir lassen~~ statt dessen	+	Nachdem <math>z</math> und <math>\bar{z}</math> in der Gleichung sind, können wir <math>z</math> (oder <math>\bar{z}</math>) nicht dirket benutzen um die Gleichung zu lösen, benutzen statt dessen

	{{Abgesetzte Formel\|\|<math>z=x+iy</math>}}		{{Abgesetzte Formel\|\|<math>z=x+iy</math>}}
Zeile 5:		Zeile 5:
	und lösen die Gleichung für den Realteil <math>x</math> und den Imaginärteil <math>y</math>.		und lösen die Gleichung für den Realteil <math>x</math> und den Imaginärteil <math>y</math>.

-	Die linke Seite der Gleichung ~~bekommt~~ dann	+	Die linke Seite der Gleichung wird dann

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 14:		Zeile 14:
	\end{align}</math>}}		\end{align}</math>}}

-	~~und~~ die Gleichung ~~ist also~~	+	also ist die Gleichung

	{{Abgesetzte Formel\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}

-	Die beiden Seiten sind gleich wenn deren Real- und Imaginärteile gleich sind, ~~und also~~ erhalten wir	+	Die beiden Seiten sind gleich, wenn deren Real- und Imaginärteile gleich sind, daher erhalten wir

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}

Martin um 10:44, 12. Mai 2009

2009-05-12T10:44:51Z

← Nächstältere Version		Version vom 10:44, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~Because the equation contains both~~ <math>z</math> ~~and~~ <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) ~~alone as an unknown~~, ~~so we are forced to set~~	+	Nachdem <math>z</math> und <math>\bar{z}</math> in der Gleichung, we cannot use <math>z</math> (oder <math>\bar{z}</math>) lösen, und wir lassen statt dessen

	{{Abgesetzte Formel\|\|<math>z=x+iy</math>}}		{{Abgesetzte Formel\|\|<math>z=x+iy</math>}}

-	~~and use the real part~~ <math>x</math> ~~and the imaginary part~~ <math>y</math> ~~as unknowns~~.	+	und lösen die Gleichung für den Realteil <math>x</math> und den Imaginärteil <math>y</math>.

-	~~With this approach, the left-hand side of the equation becomes~~	+	Die linke Seite der Gleichung bekommt dann

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 14:		Zeile 14:
	\end{align}</math>}}		\end{align}</math>}}

-	~~and the whole equation becomes~~	+	und die Gleichung ist also

	{{Abgesetzte Formel\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}

-	~~The two complex numbers on the left~~- ~~and right-hand sides are equal when both real and imaginary parts are equal~~, ~~i.e.~~	+	Die beiden Seiten sind gleich wenn deren Real- und Imaginärteile gleich sind, und also erhalten wir

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}

-	~~This gives~~ <math>x=3</math> ~~and~~ <math>y=2x-5=2\cdot 3-5=1</math>. ~~Thus, the equation has the solution~~ <math>z=3+i</math>.	+	Dies ergibt <math>x=3</math> und <math>y=2x-5=2\cdot 3-5=1</math>. Also ist die Lösung der Gleichung <math>z=3+i</math>.

-	~~A quick check shows that~~ <math>z=3+i</math> ~~satisfies the equation~~ in ~~the exercise~~,	+	Dies überprüfen wir einfach, indem wir <math>z=3+i</math> in der ursprünglichen Gleichung substituieren,

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

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

2009-03-11T10:33:55Z

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

← Nächstältere Version

Version vom 10:33, 11. Mär. 2009

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

2009-03-10T13:07:24Z

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

← Nächstältere Version		Version vom 13:07, 10. Mär. 2009
Zeile 1:		Zeile 1:
	Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set		Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set

-	{{~~Displayed math~~\|\|<math>z=x+iy</math>}}	+	{{Abgesetzte Formel\|\|<math>z=x+iy</math>}}

	and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.		and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
Zeile 7:		Zeile 7:
	With this approach, the left-hand side of the equation becomes		With this approach, the left-hand side of the equation becomes

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(1+i)(x-iy)+i(x+iy)		(1+i)(x-iy)+i(x+iy)
	&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]		&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]
Zeile 16:		Zeile 16:
	and the whole equation becomes		and the whole equation becomes

-	{{~~Displayed math~~\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}

	The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.		The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}

	This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.		This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
Zeile 26:		Zeile 26:
	A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,		A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{LHS}		\text{LHS}
	&= (1+i)\bar{z}+iz\\[5pt]		&= (1+i)\bar{z}+iz\\[5pt]

Tek um 08:30, 30. Okt. 2008

2008-10-30T08:30:05Z

← Nächstältere Version		Version vom 08:30, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}
	Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set		Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set

-		+	{{Displayed math\|\|<math>z=x+iy</math>}}
-	<math>z=x+iy</math>	+
-		+

	and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.		and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
Zeile 10:		Zeile 7:
	With this approach, the left-hand side of the equation becomes		With this approach, the left-hand side of the equation becomes

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>(1+i)(x-iy)+i(x+iy)~~</math>~~	+	(1+i)(x-iy)+i(x+iy)
-		+	&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]
-	~~<math>~~	+	&= x-iy+ix+y+ix-y\\[5pt]
-	~~\begin{align}~~ &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\	+	&= x+(2x-y)i
-	&=x-iy+ix+y+ix-y\\	+	\end{align}</math>}}
-	&=x+(2x-y)i\end{align}</math>	+
-		+

	and the whole equation becomes		and the whole equation becomes

-		+	{{Displayed math\|\|<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}
-	<math>x+(2x-y)i=3+5i</math>.	+
-		+

	The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.		The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.

-		+	{{Displayed math\|\|<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}
-	<math>\begin{~~cases~~}x=3\\2x-y=5.\end{~~cases~~}</math>	+
-		+

	This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.		This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.

-	A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise:	+	A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,
-		+
-		+
-	~~<math>\begin{align}LHS &= (1+i)\bar{z}+iz=(1+i)(3-i)+i(3+i)\\~~	+
-	~~&= 3-i+3i+1+3i-1 = 3+5i = RHS\end{align}</math>~~	+

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

Jamesjmnewton um 11:46, 23. Sep. 2008

2008-09-23T11:46:39Z

← Nächstältere Version		Version vom 11:46, 23. Sep. 2008
Zeile 11:		Zeile 11:


-	<math>(1+i)(x-iy)+i(x+iy)</math><math>	+	<math>(1+i)(x-iy)+i(x+iy)</math>
		+
		+	<math>
	\begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\		\begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\
	&=x-iy+ix+y+ix-y\\		&=x-iy+ix+y+ix-y\\

Jamesjmnewton um 11:46, 23. Sep. 2008

2008-09-23T11:46:21Z

← Nächstältere Version		Version vom 11:46, 23. Sep. 2008
Zeile 11:		Zeile 11:


-	<math>(1+i)(x-iy)+i(x+iy)	+	<math>(1+i)(x-iy)+i(x+iy)</math><math>
	\begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\		\begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\
	&=x-iy+ix+y+ix-y\\		&=x-iy+ix+y+ix-y\\
Zeile 26:		Zeile 26:


-	<math>\begin{cases}x=3\\2x-y=5\end{cases}</math>.	+	<math>\begin{cases}x=3\\2x-y=5.\end{cases}</math>

Jamesjmnewton um 11:45, 23. Sep. 2008

2008-09-23T11:45:40Z

← Nächstältere Version		Version vom 11:45, 23. Sep. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_4f-1~~(2)~~.gif]]~~ </~~center~~>	+	Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set
-	{{~~NAVCONTENT_STOP}~~}	+
-	{{~~NAVCONTENT_START}~~}	+
-	<~~center~~> ~~[[Image:3_1_4f~~-2(2)~~.gif]]~~ </~~center~~>	+	<math>z=x+iy</math>
		+
		+
		+	and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
		+
		+	With this approach, the left-hand side of the equation becomes
		+
		+
		+	<math>(1+i)(x-iy)+i(x+iy)
		+	\begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\
		+	&=x-iy+ix+y+ix-y\\
		+	&=x+(2x-y)i\end{align}</math>
		+
		+
		+	and the whole equation becomes
		+
		+
		+	<math>x+(2x-y)i=3+5i</math>.
		+
		+
		+	The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
		+
		+
		+	<math>\begin{cases}x=3\\2x-y=5\end{cases}</math>.
		+
		+
		+	This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
		+
		+	A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise:
		+
		+
		+	<math>\begin{align}LHS &= (1+i)\bar{z}+iz=(1+i)(3-i)+i(3+i)\\
		+	&= 3-i+3i+1+3i-1 = 3+5i = RHS\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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

2008-09-16T14:54:21Z

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

← Nächstältere Version

Version vom 14:54, 16. Sep. 2008

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

2008-08-21T08:06:54Z

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

← Nächstältere Version		Version vom 08:06, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_4f-1(2).gif]] </center>	+	<center> [[Image:3_1_4f-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_4f-2(2).gif]] </center>	+	<center> [[Image:3_1_4f-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}