Numere complexe si puteri
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Numere complexe si puteri
Fie \( z\in\mathbb{C},m,n\in\mathbb{N}^* \), astfel incat \( z^m=1,\ (1+z)^n=1 \). Sa se arate ca \( z^3=1 \).
n-ar fi rau sa fie bine 
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Din prima relatie
\( z=\cos \frac{2k\pi}{n}+i\sin \frac{2k\pi}{n} \)
iar din a doua relatie
\( 2\cos^2 \frac{k\pi}{n}+2\sin \frac{k\pi}{n}\cos \frac{k\pi}{n}=\cos \frac{2p\pi}{m}+i\sin \frac{2p\pi}{m} \).
Se obtine \( \cos \frac{k\pi}{n}=\pm\frac{1}{2} \), de unde n=3k sau 2n=3k, ceea ce conduce la concluzie.
\( z=\cos \frac{2k\pi}{n}+i\sin \frac{2k\pi}{n} \)
iar din a doua relatie
\( 2\cos^2 \frac{k\pi}{n}+2\sin \frac{k\pi}{n}\cos \frac{k\pi}{n}=\cos \frac{2p\pi}{m}+i\sin \frac{2p\pi}{m} \).
Se obtine \( \cos \frac{k\pi}{n}=\pm\frac{1}{2} \), de unde n=3k sau 2n=3k, ceea ce conduce la concluzie.
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Imi pare rau, dar simt nevoia sa o spun ca nu putem renunta usor la "vopsirea" unor probleme, unele dintre ele chiar simple. Se observa ca \( z\ne 1 \) si \( |z|=|z+1|=1 \) , adica \( z\overline z=(1+z)(1+\overline z)=1 \) ceea ce inseamna \( z\overline z=1 \) si \( z+\overline z=-1 \) . In concluzie, \( z^2+z+1=0 \) \( \Longrightarrow \) \( z^3=1 \) .