Lösung 3.1:2a - Versionsgeschichte

Martin um 08:47, 12. Mai 2009

2009-05-12T08:47:09Z

← Nächstältere Version		Version vom 08:47, 12. Mai 2009
Zeile 1:		Zeile 1:
-	~~A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator~~,	+	Wir berechnen den Bruch indem wir ihn mit den konjugiert komplexen Nenner erweitern,

	{{Abgesetzte Formel\|\|<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}

-	~~Then, the formula for the difference of two squares gives that the new denominator is a real number~~,	+	Durch die Binomische Formel erhalten wir den Nenner,

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

-	~~All that remains is to multiply together what is in the numerator~~,	+	Jetzt berechnen wir nur noch den Zähler,

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

Tekbot: Solution 3.1:2a moved to Lösung 3.1:2a: Robot: moved page

2009-03-11T10:30:35Z

Solution 3.1:2a moved to Lösung 3.1:2a: Robot: moved page

← Nächstältere Version

Version vom 10:30, 11. Mär. 2009

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

2009-03-10T13:05:44Z

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

← Nächstältere Version		Version vom 13:05, 10. Mär. 2009
Zeile 1:		Zeile 1:
	A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,		A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,

-	{{~~Displayed math~~\|\|<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}

	Then, the formula for the difference of two squares gives that the new denominator is a real number,		Then, the formula for the difference of two squares gives that the new denominator is a real number,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}		\frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}
	&= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt]		&= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt]
Zeile 15:		Zeile 15:
	All that remains is to multiply together what is in the numerator,		All that remains is to multiply together what is in the numerator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{(3-2i)(1-i)}{2}		\frac{(3-2i)(1-i)}{2}
	&= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt]		&= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt]

Tek um 15:15, 29. Okt. 2008

2008-10-29T15:15:19Z

← Nächstältere Version		Version vom 15:15, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,
-	A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator:	+

-	<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\frac{1-i}{1-i}.</math>	+	{{Displayed math\|\|<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.}</math>}}
-	~~{{NAVCONTENT_STEP~~}}	+
-	~~Then, the conjugate rule gives that the new denominator is a real number~~	+

-	~~<math>\begin{align}\frac{3-2i}{1+i}\frac{1-i}{1-i}&=\frac{(3-2i)(1-i)}{(1+i)(1-i)}\\~~	+	Then, the formula for the difference of two squares gives that the new denominator is a real number,
-	~~&=\frac{(3-2i)(1-i)}{1^2-i^2}\\~~	+
-	~~&=\frac{(3-2i)(1-i)}{1+1}\\~~	+
-	~~&=\frac{(3-2i)(1-i)}{2}\end{align}</math>~~	+
-	~~{{NAVCONTENT_STEP}}~~	+
-	~~All~~ that ~~remains is to multiply together what is in~~ the ~~numerator:~~	+

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

-	~~<math>\begin{align}\frac{(3-2i)(1-i)}{2}&=\frac{3\cdot 1 -3\cdot i - 1\cdot 2i -2i\cdot(-i)}{2}\\~~	+	All that remains is to multiply together what is in the numerator,
-	~~&=\frac{3-3i-2i+2i^2}{2}\\~~	+
-	~~&=\frac{3-(3+2)i+2(-1)}{2}\\~~	+
-	~~&=\frac{1-5i}{2}\\~~	+
-	~~&=\frac{1}{2}-\frac{5}{2}i.\end{align}</math>~~	+

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

Jamesjmnewton um 09:39, 23. Sep. 2008

2008-09-23T09:39:03Z

← Nächstältere Version		Version vom 09:39, 23. Sep. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_2a~~-1~~(2)~~.~~gif]]~~ </~~center~~>	+	A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator:
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\frac{3-2i}{1+i} = \frac{3-2i}{1+i}\frac{1-i}{1-i}.</math>
-	<~~center~~> ~~[[Image~~:~~3_1_2a~~-2(2).~~gif]]~~ </~~center~~>	+	{{NAVCONTENT_STEP}}
		+	Then, the conjugate rule gives that the new denominator is a real number
		+
		+	<math>\begin{align}\frac{3-2i}{1+i}\frac{1-i}{1-i}&=\frac{(3-2i)(1-i)}{(1+i)(1-i)}\\
		+	&=\frac{(3-2i)(1-i)}{1^2-i^2}\\
		+	&=\frac{(3-2i)(1-i)}{1+1}\\
		+	&=\frac{(3-2i)(1-i)}{2}\end{align}</math>
		+	{{NAVCONTENT_STEP}}
		+	All that remains is to multiply together what is in the numerator:
		+
		+
		+	<math>\begin{align}\frac{(3-2i)(1-i)}{2}&=\frac{3\cdot 1 -3\cdot i - 1\cdot 2i -2i\cdot(-i)}{2}\\
		+	&=\frac{3-3i-2i+2i^2}{2}\\
		+	&=\frac{3-(3+2)i+2(-1)}{2}\\
		+	&=\frac{1-5i}{2}\\
		+	&=\frac{1}{2}-\frac{5}{2}i.\end{align}</math>
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot: Lösning 3.1:2a moved to Solution 3.1:2a: Robot: moved page

2008-09-16T14:51:01Z

Lösning 3.1:2a moved to Solution 3.1:2a: Robot: moved page

← Nächstältere Version

Version vom 14:51, 16. Sep. 2008

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

2008-08-21T08:05:14Z

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

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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_2a-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_1_2a-2(2).gif {{NAVCONTENT_STOP}}

2008-04-07T10:43:41Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_2a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_1_2a-2(2).gif </center> {{NAVCONTENT_STOP}}

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