Fie \( z_{1}, z_{2}, z_{3}\in\mathbb{C} \) numere complexe astfel incat
\( |z_{1}|=|z_{2}|=|z_{3}|=1 \) si \( z_{1}^{3}+z_{2}^{3}+z_{3}^{3}+z_{1}z_{2}z_{3}=0 \). Sa se arate ca \( |z_{1}+z_{2}+z_{3}|\in\{1,2\} \).
Trei numere complexe de modul 1 au suma de modulul 1 sau 2
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Trei numere complexe de modul 1 au suma de modulul 1 sau 2
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.
Folosim identitatea :
\( z_1^3+z_2^3+z_3^3-3z_1z_2z_3=(z_1+z_2+z_3)(z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1) \)
Din relatia din ipoteza avem ca:
\( (z_1+z_2+z_3)(z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1)+4z_1z_2z_3=(z_1+z_2+z_3)^3-3(z_1z_2+z_2z_3+z_3z_1)(z_1+z_2+z_3)+4z_1z_2z_3=0 \)
Dar avand in vedere ca \( \left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1 \) obtinem:
\( (z_1z_2+z_2z_3+z_3z_1)(z_1+z_2+z_3)=z_1z_2z_3(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})(z_1+z_2+z_3)=z_1z_2z_3(z_1+z_2+z_3)(\bar{z_1+z_2+z_3})=z_1z_2z_3\left|z_1+z_2+z_3\right|^2 \)
Deci
\( (z_1+z_2+z_3)^3=z_1z_2z_3(3\left|z_1+z_2+z_3\right|^2-4) \)
Notam \( \left| z_1+z_2+z_3 \right|=Z \).Trecand la modul obtinem ecuatia :
\( Z^3=\left|3Z^2-4\right| \)
Daca \( \left|3Z^2-4\right|=3Z^2-4 \) adica pentru \( Z^2\geq \frac{4}{3} \) avem:\( Z^3-3Z^2+4=(Z-2)^2(Z+1)=0 \)
Iar daca \( \left|3Z^2-4\right|=4-3Z^2 \) adica pentru \( Z^2 \leq \frac{4}{3} \) avem \( Z^3+3Z^2-4=(Z-1)(Z+2)^2=0 \)
Din cele arate , obtinem \( Z=\left|z_1+z_2+z_3\right| \in \left{1,2\right} \)
\( z_1^3+z_2^3+z_3^3-3z_1z_2z_3=(z_1+z_2+z_3)(z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1) \)
Din relatia din ipoteza avem ca:
\( (z_1+z_2+z_3)(z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1)+4z_1z_2z_3=(z_1+z_2+z_3)^3-3(z_1z_2+z_2z_3+z_3z_1)(z_1+z_2+z_3)+4z_1z_2z_3=0 \)
Dar avand in vedere ca \( \left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1 \) obtinem:
\( (z_1z_2+z_2z_3+z_3z_1)(z_1+z_2+z_3)=z_1z_2z_3(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})(z_1+z_2+z_3)=z_1z_2z_3(z_1+z_2+z_3)(\bar{z_1+z_2+z_3})=z_1z_2z_3\left|z_1+z_2+z_3\right|^2 \)
Deci
\( (z_1+z_2+z_3)^3=z_1z_2z_3(3\left|z_1+z_2+z_3\right|^2-4) \)
Notam \( \left| z_1+z_2+z_3 \right|=Z \).Trecand la modul obtinem ecuatia :
\( Z^3=\left|3Z^2-4\right| \)
Daca \( \left|3Z^2-4\right|=3Z^2-4 \) adica pentru \( Z^2\geq \frac{4}{3} \) avem:\( Z^3-3Z^2+4=(Z-2)^2(Z+1)=0 \)
Iar daca \( \left|3Z^2-4\right|=4-3Z^2 \) adica pentru \( Z^2 \leq \frac{4}{3} \) avem \( Z^3+3Z^2-4=(Z-1)(Z+2)^2=0 \)
Din cele arate , obtinem \( Z=\left|z_1+z_2+z_3\right| \in \left{1,2\right} \)
Vrajitoarea Andrei