Processing Math: Done
Antwort 3.3:6
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
(Ny sida: Lösningar: <math>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\,...) |
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| Zeile 1: | Zeile 1: | ||
Lösningar: | Lösningar: | ||
<math>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.</math> | <math>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.</math> | ||
| + | Uttryck: | ||
| + | <math>\displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1</math> | ||
Version vom 13:13, 7. Apr. 2008
Lösningar:
42
cos
8+isin8
42cos89+isin89=
21
22+2+i2122−2−2122+2−i2122−2
8=2−1
