Tek: Ändrat svaret till c-uppgiften

Tek — Mon, 21 Dec 2009 12:07:02 GMT

Ändrat svaret till c-uppgiften

← Äldre version		Versionen från 21 december 2009 kl. 12.07
Rad 6:		Rad 6:
	\|-		\|-
	\|c)		\|c)
-	\|width="50%"\| <math>z= \left\{\begin{matrix} -1 \\ \~~phantom{~~-}3 \\ \end{matrix}\right. </math>	+	\|width="50%"\| <math>z= \left\{\begin{matrix} -1+i\sqrt{2} \\ -1-i\sqrt{2} \\ \end{matrix}\right. </math>
	\|d)		\|d)
	\|width="50%"\| <math>z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right.</math>		\|width="50%"\| <math>z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right.</math>
	\|}		\|}

Oskar: Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%"| $z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right.$ |b) |width="50%"| $z = \left...$

Oskar — Mon, 07 Apr 2008 13:03:11 GMT

Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%"|<math>z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right.</math> |b) |width="50%"| <math>z = \left...

Ny sida

{| width="100%" cellspacing="10px"
|a)
|width="50%"|<math>z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right.</math>
|b)
|width="50%"| <math>z = \left\{\begin{matrix} 2+i \\ 2-i \\ \end{matrix}\right.</math>
|-
|c)
|width="50%"| <math>z= \left\{\begin{matrix} -1 \\ \phantom{-}3 \\ \end{matrix}\right. </math>
|d)
|width="50%"| <math>z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right.</math>
|}

Svar 3.3:4 - Versionshistorik

Tek: Ändrat svaret till c-uppgiften

Oskar: Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%"|z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right. |b) |width="50%"| z = \left...

Oskar: Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%"| $z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right.$ |b) |width="50%"| $z = \left...$