Svar 3.3:4

Förberedande kurs i matematik 2

(Skillnad mellan versioner)
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(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...)
Nuvarande version (31 juli 2009 kl. 10.27) (redigera) (ogör)
(Korrigerat svaret i c)
 
Rad 6: Rad 6:
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|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>
|}
|}

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.