Lösung 3.1:3 - Versionsgeschichte

Martin um 09:11, 12. Mai 2009

2009-05-12T09:11:37Z

← Nächstältere Version		Version vom 09:11, 12. Mai 2009
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-	~~In order to be able to see the expression's real and imaginary parts directly~~, ~~we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the denominator~~,	+	Wir müssen den Ausdruck in Real- und Imaginärteil aufteilen, um zu sehen wann der Ausdruck rein imaginär ist. Wir erweitern dazu den Bruch mit den konjugiert komplexen Nenner,

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

-	The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.

		+	Der Realteil des Ausdruckes ist null wenn <math>6+a=0</math>, also wenn <math>a=-6</math>.

-	~~Note~~: ~~Think about how to solve the problem if~~ <math>a</math> ~~is not a real number.~~	+
		+	Hinweis: Wie würden wir das Problem lösen, wenn <math>a</math> nicht reell ist?

2009-03-11T10:31:55Z

Solution 3.1:3 moved to Lösung 3.1:3: Robot: moved page

← Nächstältere Version

Version vom 10:31, 11. Mär. 2009

2009-03-10T13:06:24Z

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

← Nächstältere Version		Version vom 13:06, 10. Mär. 2009
Zeile 1:		Zeile 1:
	In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the denominator,		In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the denominator,

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

2008-10-30T07:11:13Z

← Nächstältere Version		Version vom 07:11, 30. Okt. 2008
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-	~~{{NAVCONTENT_START}}~~	+	In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the denominator,
-	In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the ~~numerator:~~	+
-		+
-		+
-	~~<math>\begin{align}\frac{3+i}{2+ai}&=\frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\~~	+
-	~~&=\frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\~~	+
-	~~&=\frac{6+a+(2-3a)i}{4+a^2}\\~~	+
-	~~&=\frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}i.\end{align}</math>~~	+

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

	The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.		The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.

-	NOTE: Think about how to solve the problem if a is not a real number.

-	~~{{NAVCONTENT_STOP}}~~	+	Note: Think about how to solve the problem if <math>a</math> is not a real number.

2008-09-25T16:13:01Z

← Nächstältere Version		Version vom 16:13, 25. Sep. 2008
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	The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.		The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.

-	NOTE: Think about how to solve the problem is a is not a real number.	+	NOTE: Think about how to solve the problem if a is not a real number.

	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-09-23T10:29:46Z

← Nächstältere Version		Version vom 10:29, 23. Sep. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<~~center~~> ~~[[Image:3_1_3~~.~~gif]]~~ </~~center~~>	+	In order to be able to see the expression's real and imaginary parts directly, we treat it as an ordinary quotient of two complex numbers and multiply top and bottom by the complex conjugate of the numerator:
		+
		+
		+	<math>\begin{align}\frac{3+i}{2+ai}&=\frac{(3+i)(2-ai)}{(2+ai)(2-ai)}\\
		+	&=\frac{3\cdot 2-3\cdot ai +i\cdot 2-ai^2}{2^2-(ai)^2}\\
		+	&=\frac{6+a+(2-3a)i}{4+a^2}\\
		+	&=\frac{6+a}{4+a^2}+\frac{2-3a}{4+a^2}i.\end{align}</math>
		+
		+
		+	The expression has real part equal to zero when <math>6+a=0</math>, i.e. <math>a=-6</math>.
		+
		+	NOTE: Think about how to solve the problem is a is not a real number.
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

2008-09-16T14:52:21Z

Lösning 3.1:3 moved to Solution 3.1:3: Robot: moved page

← Nächstältere Version

Version vom 14:52, 16. Sep. 2008

2008-08-21T08:05:54Z

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

2008-04-07T10:51:10Z

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