Matematyka.pl

\(\displaystyle{ z=(\sqrt{5} -\iota \sqrt{5})^5}\)

moduł liczby zespolonej = \(\displaystyle{ \sqrt{10}}\)

\(\displaystyle{ \sin\phi= -\frac{\sqrt{2}}{2}}\)
\(\displaystyle{ \cos\phi=\frac{\sqrt{2}}{2}}\)

kąty leżą w IV ćwiartce

\(\displaystyle{ z=\sqrt{10}(\cos\frac{7\pi}{4} + \iota\sin\frac{7\pi}{4})}\)

\(\displaystyle{ 100\sqrt{10}(\cos(8\pi+\frac{3\pi}{4})+\iota\sin(8\pi+\frac{3\pi}{4}))}\)
W jakim celu do 3/4 pi dodajemy te 8 pi?NIer ozumiem tego troche...

Na koniec zostaje \(\displaystyle{ 100\sqrt{10}(\cos(\pi-\frac{\pi}{4})+\iota\sin(\pi-\frac{\pi}{4})}\)

[ Dodano: 21 Października 2007, 14:50 ]
a no i skad sie wzielo \(\displaystyle{ \frac{7\pi}{4}}\)

\(\displaystyle{ \cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}}\)
\(\displaystyle{ \sin\frac{7\pi}{4}=\frac{-\sqrt{2}}{2}}\)
\(\displaystyle{ z^{5}=100\sqrt{10}(cis(\frac{7\pi}{4}))^{5}}\)
\(\displaystyle{ z^{5}=100\sqrt{10}(cis(\frac{35\pi}{4}))}\)

korzystasz z tego ze \(\displaystyle{ \cos\alpha=\cos(\alpha+2k\pi)}\) i to samo dla sinusa
\(\displaystyle{ z^{5}=100\sqrt{10}(cis(\frac{3\pi}{4}))}\)
\(\displaystyle{ z^{5}=100\sqrt{10}(\frac{-\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})}\)

ok dzieki wielkie