Logika, zadanie nr 699
ostatnie wiadomości | regulamin | latex
| Autor | Zadanie / Rozwiązanie |
| 323232 postów: 22 |  2012-11-25 22:48:09 |
| abcdefgh postów: 1255 | 2013-07-02 20:37:21 a) $(1-i)^{12}$ $|z|=\sqrt{1^2+1^2}=\sqrt{2}$ $sin\phi=\frac{-1}{\sqrt{2}} \ \ \ cos\phi=\frac{1}{\sqrt{2}}$ $\phi =\frac{7\pi}{4}$ $z^{12}=\sqrt{2}^{12}(isin(12*\frac{7\pi}{4})+cos(12*\frac{7\pi}{4}))=1024*(isin21\pi+cos21\pi)$ b)$1+\sqrt{3}i^{8}=1-\sqrt{3}i^{6}=1+\sqrt{3}i^{4}=1-\sqrt{3}i^{2}=1+\sqrt{3}=$ $c)\ (2*\sqrt{3}-2i)^{30}$ $|z|=\sqrt{4+12}=4$ $sin\phi=\frac{-1}{2} \ \ \ cos\phi=\frac{\sqrt{3}}{2}$ $\phi=\frac{11\pi}{6}$ $z^{30}=4^{30}*(cos(\frac{11\pi}{6}*30)+isin(30*\frac{11\pi}{6}))=4^{30}*(cos55\pi+isin55\pi)$ d)$\sqrt{5-12i}$ $|z|=\sqrt{25+144}=13$ $sin\phi=\frac{-12}{13} \ \ \ cos\phi=\frac{5}{13}$ $\phi=\frac{67\pi}{180}$ $z_{0}=\sqrt{n}{13}*[cos(\frac{67\pi}{2\pi})+isin(\frac{67}{\pi})]$ $z_{1}=\sqrt{n}{13}*[cos(\frac{67}{\pi}+\pi)+isin(\frac{67}{\pi}+\pi)]$ e) $\sqrt[3]{-1+i}$ $|z|=\sqrt{2}$ $sin\phi=\frac{1}{\sqrt{2}}$ $cos\phi=\frac{-1}{\sqrt{2}}$ $\phi=\frac{3\pi}{4}$ $z_{0}=\frac{6}{3}[cos(\frac{3\pi}{12})+isin(\frac{3\pi}{12})]$ $z_{1}=\frac{6}{3}[cos(\frac{11\pi}{12})+isin(\frac{11\pi}{12})]$ $z_{2}=\frac{6}{3}[cos(\frac{19\pi}{12})+isin(\frac{19\pi}{12})]$ |
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