łatwe
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Sir George
- Użytkownik
- Posty: 1145
- Rejestracja: 27 kwie 2006, o 10:19
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łatwe
...czyli
\(\displaystyle{ z^4\ =\ -i\ =\ \cos\frac32\pi\,+\,i\,\sin\frac32\pi}\)
Zatem
\(\displaystyle{ z_1\ =\ \cos\frac38\pi\,+\,i\,\sin\frac38\pi\ =\ \frac{\sqrt{2-\sqrt{2}}}{2}\,+\,i\,\frac{\sqrt{2+\sqrt{2}}}{2} \\ z_2\ =\ \cos\frac78\pi\,+\,i\,\sin\frac78\pi\ =\ -\frac{\sqrt{2+\sqrt{2}}}{2}\,+\,i\,\frac{\sqrt{2-\sqrt{2}}}{2} \\ z_3\ =\ \cos\frac{11}8\pi\,+\,i\,\sin\frac{11}8\pi\ =\ -\frac{\sqrt{2-\sqrt{2}}}{2}\,-\,i\,\frac{\sqrt{2+\sqrt{2}}}{2} \\ z_4\ =\ \cos\frac{15}8\pi\,+\,i\,\sin\frac{15}8\pi\ =\ \frac{\sqrt{2+\sqrt{2}}}{2}\,-\,i\,\frac{\sqrt{2-\sqrt{2}}}{2}}\)
\(\displaystyle{ z^4\ =\ -i\ =\ \cos\frac32\pi\,+\,i\,\sin\frac32\pi}\)
Zatem
\(\displaystyle{ z_1\ =\ \cos\frac38\pi\,+\,i\,\sin\frac38\pi\ =\ \frac{\sqrt{2-\sqrt{2}}}{2}\,+\,i\,\frac{\sqrt{2+\sqrt{2}}}{2} \\ z_2\ =\ \cos\frac78\pi\,+\,i\,\sin\frac78\pi\ =\ -\frac{\sqrt{2+\sqrt{2}}}{2}\,+\,i\,\frac{\sqrt{2-\sqrt{2}}}{2} \\ z_3\ =\ \cos\frac{11}8\pi\,+\,i\,\sin\frac{11}8\pi\ =\ -\frac{\sqrt{2-\sqrt{2}}}{2}\,-\,i\,\frac{\sqrt{2+\sqrt{2}}}{2} \\ z_4\ =\ \cos\frac{15}8\pi\,+\,i\,\sin\frac{15}8\pi\ =\ \frac{\sqrt{2+\sqrt{2}}}{2}\,-\,i\,\frac{\sqrt{2-\sqrt{2}}}{2}}\)