Svar 3.3:4
Förberedande kurs i matematik 2
(Skillnad mellan versioner)
(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...) |
(Ändrat svaret till c-uppgiften) |
||
| Rad 6: | Rad 6: | ||
|- | |- | ||
|c) | |c) | ||
| - | |width="50%"| <math>z= \left\{\begin{matrix} -1 \\ \ | + | |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> | ||
|} | |} | ||
Nuvarande version
| a) | \displaystyle z= \left\{\begin{matrix} \phantom{-}(1+i)/\sqrt{2}\\ -(1+i)/\sqrt{2}\\ \end{matrix}\right. | b) | \displaystyle z = \left\{\begin{matrix} 2+i \\ 2-i \\ \end{matrix}\right. |
| c) | \displaystyle z= \left\{\begin{matrix} -1+i\sqrt{2} \\ -1-i\sqrt{2} \\ \end{matrix}\right. | d) | \displaystyle z= \left\{\begin{matrix} (1+i\sqrt{15})/4\\ (1-i\sqrt{15})/4 \end{matrix}\right. |
