Antwort 3.3:6
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Robot: Automated text replacement (-Solution +Lösung)) |
K (Answer 3.3:6 moved to Antwort 3.3:6: Robot: moved page) |
Version vom 09:14, 11. Mär. 2009
| Lösungs: | \displaystyle z= \left\{\eqalign{&\textstyle\sqrt[\scriptstyle 4]{2}\bigl(\cos\frac{\pi}{8}+i\,\sin\frac{\pi}{8}\bigr)\cr &\textstyle\sqrt[\scriptstyle 4]{2}\bigl(\cos\frac{9\pi}{8}+i\,\sin\frac{9\pi}{8}\bigr)}\right. = \left\{\eqalign{&\textstyle\phantom{-}{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}+2}\vphantom{2^{2^{\scriptstyle 2}}}}+i\,{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}-2}\vphantom{2^{2^{\scriptstyle 2}}}}\cr &\textstyle -{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}+2}\vphantom{2^{2^{\scriptstyle 2}}}}-i\,{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}-2}\vphantom{2^{2^{\scriptstyle 2}}}}}\right. |
| Expression: | \displaystyle \displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1 |
