Antwort 3.3:6

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(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\,...)
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>
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Uttryck:
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<math>\displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1</math>

Version vom 13:13, 7. Apr. 2008

Lösningar: \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. Uttryck: \displaystyle \displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1

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