Antwort 3.3:2 - Versionsgeschichte

Dima um 12:05, 5. Sep. 2009

2009-09-05T12:05:43Z

← Nächstältere Version		Version vom 12:05, 5. Sep. 2009
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	\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>		\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>
	\|c)		\|c)
-	\|width="33%"\| <math>~~\displaystyle~~ z=2^{1/10}~~\exp\Bigl~~(\frac{\pi i}{4}+\frac{2n\pi i}{5}\~~Bigr~~)~~</math> für <math>~~\ ~~n=0,~~1,2~~,3,4~~</math>	+	\|width="33%"\|<math>z= \left\{\begin{matrix} 2^{\frac{1}{10}}e^{i(\frac{\pi}{4})} \\ 2^{\frac{1}{10}}e^{i(\frac{13\pi}{20})} \\ 2^{\frac{1}{10}}e^{i(\frac{21\pi}{20})} \\ 2^{\frac{1}{10}}e^{i(\frac{29\pi}{20})}\\ 2^{\frac{1}{10}}e^{i(\frac{37\pi}{20})} \\ \end{matrix}\right.</math>
	\|-		\|-
	\|d)		\|d)

Moni um 17:56, 3. Sep. 2009

2009-09-03T17:56:27Z

← Nächstältere Version		Version vom 17:56, 3. Sep. 2009
Zeile 5:		Zeile 5:
	\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>		\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>
	\|c)		\|c)
-	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)</math> für <math>\ k=0,1,2,3,4</math>	+	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2n\pi i}{5}\Bigr)</math> für <math>\ n=0,1,2,3,4</math>
	\|-		\|-
	\|d)		\|d)

Martin um 14:30, 18. Mai 2009

2009-05-18T14:30:14Z

← Nächstältere Version		Version vom 14:30, 18. Mai 2009
Zeile 5:		Zeile 5:
	\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>		\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>
	\|c)		\|c)
-	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)</math> ~~for~~ <math>\ k=0,1,2,3,4</math>	+	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)</math> für <math>\ k=0,1,2,3,4</math>
	\|-		\|-
	\|d)		\|d)

Tekbot: Answer 3.3:2 moved to Antwort 3.3:2: Robot: moved page

2009-03-11T09:13:11Z

Answer 3.3:2 moved to Antwort 3.3:2: Robot: moved page

← Nächstältere Version

Version vom 09:13, 11. Mär. 2009

Tekbot: Svar 3.3:2 moved to Answer 3.3:2: Robot: moved page

2008-09-16T14:07:57Z

Svar 3.3:2 moved to Answer 3.3:2: Robot: moved page

← Nächstältere Version

Version vom 14:07, 16. Sep. 2008

Tek um 14:30, 5. Sep. 2008

2008-09-05T14:30:04Z

← Nächstältere Version		Version vom 14:30, 5. Sep. 2008
Zeile 5:		Zeile 5:
	\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>		\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>
	\|c)		\|c)
-	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)</math> ~~för~~ <math>\ k=0,1,2,3,4</math>	+	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)</math> for <math>\ k=0,1,2,3,4</math>
	\|-		\|-
	\|d)		\|d)

Oskar um 12:43, 7. Apr. 2008

2008-04-07T12:43:05Z

← Nächstältere Version		Version vom 12:43, 7. Apr. 2008
Zeile 5:		Zeile 5:
	\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>		\|width="33%"\| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>
	\|c)		\|c)
-	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)$ för $\ k=0,1,2,3,4</math>	+	\|width="33%"\| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)</math> för <math>\ k=0,1,2,3,4</math>
	\|-		\|-
	\|d)		\|d)

Oskar: Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%"| $z= \left\{\begin{matrix} \phantom{-}1 \\ -1 \\ \phantom{-}i \\ -i\\ \end{matrix}\right.$ |b) |width="33%"| $z = \left\{\...$

2008-04-07T12:42:23Z

Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%"|<math>z= \left\{\begin{matrix} \phantom{-}1 \\ -1 \\ \phantom{-}i \\ -i\\ \end{matrix}\right.</math> |b) |width="33%"| <math>z = \left\{\...

Neue Seite

{| width="100%" cellspacing="10px"
|a)
|width="33%"|<math>z= \left\{\begin{matrix} \phantom{-}1 \\ -1 \\ \phantom{-}i \\ -i\\ \end{matrix}\right.</math>
|b)
|width="33%"| <math>z = \left\{\begin{matrix} \frac{1}{2}+\frac{1}{2}\sqrt{3}\,i \\ -1\phantom{{}-\frac{1}{2}\sqrt{3}\,i} \\ \frac{1}{2}-\frac{1}{2}\sqrt{3}\,i \\ \end{matrix}\right.</math>
|c)
|width="33%"| <math>\displaystyle z=2^{1/10}\exp\Bigl(\frac{\pi i}{4}+\frac{2k\pi i}{5}\Bigr)$ för $\ k=0,1,2,3,4</math>
|-
|d)
|width="33%"| <math>z= \left\{\begin{matrix} 2+i \\ 2-i \\ \phantom{-}i \\ -i\\ \end{matrix}\right.</math>
|e)
|width="33%"| <math>z = \left\{\begin{matrix} \phantom{-}1 \\ -1 \end{matrix}\right.</math>
|}