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Daniel za post:
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od Daniel » Subota, 27. Septembar 2014, 20:34
Pa, to je to. Sad samo uvrštavaš svaku od te četiri vrednosti za [inlmath]k[/inlmath] i dobićeš [inlmath]z_1[/inlmath], [inlmath]z_2[/inlmath], [inlmath]z_3[/inlmath] i [inlmath]z_4[/inlmath]:
[dispmath]\begin{array}{ll}
k=0:\qquad & \begin{array}{l}
z_1=\sqrt 2\left(\cos\frac{\pi+2\cdot 0\cdot\pi}{4}+i\sin\frac{\pi+2\cdot 0\cdot\pi}{4}\right)=\\
=\sqrt 2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)=\sqrt 2\left(\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right)=1+i
\end{array}\\[2em]
k=1:\qquad & \begin{array}{l}
z_2=\sqrt 2\left(\cos\frac{\pi+2\cdot 1\cdot\pi}{4}+i\sin\frac{\pi+2\cdot 1\cdot\pi}{4}\right)=\\
=\sqrt 2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)=\sqrt 2\left(-\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\right)=-1+i
\end{array}
\end{array}[/dispmath]
Isto tako uradi i za [inlmath]k=2[/inlmath] i za [inlmath]k=3[/inlmath]...
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain