Antwort 3.3:4

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Answer 3.3:4 moved to Antwort 3.3:4: Robot: moved page)
Aktuelle Version (11:25, 31. Jul. 2009) (bearbeiten) (rückgängig)
(Changed answer to 3.3:4c)
 
Zeile 6: Zeile 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>
|}
|}

Aktuelle 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.