|
|
|
| |
|
|
|
Antwort 3.3:6
Aus Online Mathematik Brückenkurs 2
| Lösungen:
| \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.
|
| Ausdruck:
| \displaystyle \displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1
|
|
|
|
|
 |
|