Fie \( z \) un numar complex pentru care \( |z| \), \( |z^2-2z+2| \) si \( |z^2-3z+3| \) sunt cel mult \( 1 \). Demonstrati ca \( z=1 \).
Virgil Nicula, Lista scurta ONM 2007
Numere complexe de modul cel mult 1, atunci z=1
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Numere complexe de modul cel mult 1, atunci z=1
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
-
opincariumihai
- Thales
- Posts: 134
- Joined: Sat May 09, 2009 7:45 pm
- Location: BRAD
Re: Numere complexe de modul cel mult 1, atunci z=1
Sa observam ca daca \( |z|\leq1 \) si \( |z^2-3z+3|\leq1 \), atunci \( |z^2-2z+2|\leq1 \). Intr-adevar, \( 3\geq2|z^2-3z+3|+|z^2|\geq|3z^2-6z+6|=3|z^2-2z+2| \).
In concluzie conditia \( |z^2-2z+2|\leq1 \) este in plus. As dori enuntul corectat al problemei.
In concluzie conditia \( |z^2-2z+2|\leq1 \) este in plus. As dori enuntul corectat al problemei.
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Notam \( z=r(\cos \phi+i\sin \phi) \) , \( 0<r\le1 \)
si relatia din enunt (\( |z^2-3z+3|\le 1 \)) se transcrie
\( (r^2\cos 2\phi-3r\cos \phi+3)^2+(r^2\sin 2\phi-3r\sin \phi)^2\le1 \)
sau
\( 12r^2(\cos\phi-\frac{r^2+3}{4r})^2+\frac{(r^2-1)(r^2-5)}{48r^2}\le 0 \)
de unde \( r=1 \) , \( \cos\phi=1 \) , \( \sin\phi=0 \), deci \( z=1 \).
si relatia din enunt (\( |z^2-3z+3|\le 1 \)) se transcrie
\( (r^2\cos 2\phi-3r\cos \phi+3)^2+(r^2\sin 2\phi-3r\sin \phi)^2\le1 \)
sau
\( 12r^2(\cos\phi-\frac{r^2+3}{4r})^2+\frac{(r^2-1)(r^2-5)}{48r^2}\le 0 \)
de unde \( r=1 \) , \( \cos\phi=1 \) , \( \sin\phi=0 \), deci \( z=1 \).