Nr. complexe

[andy crisan]

Fie \( z_1,z_2,z_3\in\mathbb C \) ai \( |z_1|=|z_2|=|z_3| \) si \( z_1+z_2=z_3 \). Calculati \( z_1^{2009}+z_2^{2009}+z_3^{2009} \).

[mihai++]

Conjugam \( z_1+z_2=z_3 \) si obtinem \( z_1z_2=z_3^2 \) si de aici \( z_1^2+z_2^2=-z_1z_2=-z_3^2 \).
Acum incercam sa vedem ce valoare ia \( z_1^k+z_2^k \):
\( (z_1^2+z_2^2)(z_1+z_2)=-z_3^2(z_1+z_2)\Leftrightarrow z_1^3+z_2^3+z_1z_2(z_1+z_2)=-z_3^3\Leftrightarrow z_1^3+z_2^3=-2z_3^3 \) si continuand sa inmultim cu \( z_1+z_2(=z_3) \) obtinem:
\( z_1^4+z_2^4=-z_3^4\\z_1^5+z_2^5=z_3^5\\z_1^6+z_2^6=2z_3^6\\z_1^7+z_2^7=z_3^7\\z_1^8+z_2^8=-z_3^8\\z_1^9+z_2^9=-2z_3^9\dots \) si de aici se observa periodicitatea cu 6 a coeficientilor din fata lui \( z_3^k \):
\( \overline{z_1^{6k}+z_2^{6k}=2z_3^{6k}\\z_1^{6k+1}+z_2^{6k+1}=z_3^{6k+1}\\z_1^{6k+2}+z_2^{6k+2}=-z_3^{6k+2}\\z_1^{6k+3}+z_2^{6k+3}=-2z_3^{6k+3}\\z_1^{6k+4}+z_2^{6k+4}=-z_3^{6k+4}\\z_1^{6k+5}+z_2^{6k+5}=z_3^{6k+5}} \). Si deci \( \underline{\overline{\left| z_1^{2009}+z_2^{2009}+z_3^{2009}=z_1^{M_6+5}+z_2^{M_6+5}+z_3^{2009}=2z_3^{2009}\right|} \).
Off topic : problema suna mult mai frumos cu 2008 caci \( z_1^{2008}+z_2^{2008}+z_3^{2008}=0 \).

[andy crisan]

Am gasit o solutie mai frumoasa pt problema asta. Fie \( \alpha,\beta \in\mathbb{C} \) a. i. \( z_2=\alpha z_1,z_3=\beta z_2=\alpha\beta z_1 \) si, trecand la module, \( \Rightarrow |\alpha|=|\beta|=1 \) si inlocuind in ecuatie \( \Rightarrow 1+\alpha=\alpha\beta \Rightarrow \beta=\frac{1+\alpha}{\alpha}\Rightarrow |\alpha+1|=1 \)din \( |1+\alpha|^2+|1-\alpha|^2=2\Rightarrow|1-\alpha|=1 \).Se rezolva sistemul\( \left{|1+\alpha|=1\\|1-\alpha|=1 si obtinem ca 1+\alpha+\alpha^2=0, \alpha^3=1,\alpha\neq1 \)si inlocuind \( \alpha \) si determinand \( \beta \), vom obtine in final ca \( z_1^{2009}+z_2^{2009} \)\( z_3^{2009}=z_1^{2009}(1+\alpha^{2009}+(-\alpha)^{2009})=z_1^{2009} \)