Numere complexe

[DrAGos Calinescu]

Fie \( z_1,z_2,z_3\in\mathbb{C} \) astfel incat \( |z_1|=|z_2|=|z_3|=1 \) si \( \frac{z_1^2}{z_2z_3}+\frac{z_2^2}{z_3z_1}+\frac{z_3^2}{z_1z_2}=-1 \)
Calculati \( S=|z_1+z_2+z_3| \)

[Mateescu Constantin]

Fie \( a\ ,\ b\ ,\ c\ \in\ \mathbb{C} \) astfel incat \( |a|=|b|=|c|=1 \) si \( \frac{a^2}{bc}\ +\ \frac{b^2}{ca}\ +\ \frac{c^2}{ab}=-1 \) . Calculati \( S=|a+b+c| \) .

Solutie :

Din identitatea valabila in \( \mathbb{C}\ :\ \underline{\overline{\left\|\ a^3+b^3+c^3-3abc=(a+b+c)\left\[(a+b+c)^2-3(ab+bc+ca)\right\]\ \right\|}} \) ,

relatia \( \frac{a^2}{bc}\ +\ \frac{b^2}{ca}\ +\ \frac{c^2}{ab}=-1 \) se scrie : \( a^3+b^3+c^3=-abc\ \Longleftrightarrow\ (a+b+c)[(a+b+c)^2-3(ab+bc+ca)]=-4abc \) .

\( \Longleftrightarrow\ (a+b+c)^3=3(a+b+c)(ab+bc+ca)-4abc\ (\ast) \) . Din \( |a|=|b|=|c|=1\ \Longrightarrow\ ab+bc+ca=\frac{1}{\overline{ab}}+\frac{1}{\overline{bc}}+\frac{1}{\overline{ca}} \)

\( \Longleftrightarrow\ ab+bc+ca=\frac{\overline{a+b+c}}{\overline{abc}} \) , deci relatia \( (\ast) \) devine : \( \overline{abc}(a+b+c)^3=3|a+b+c|^2-4|abc|^2=3S^2-4 \)

Trecand la module obtinem : \( S^3=|3S^2-4| \) . Dupa rezolvarea ecuatiei gasim \( \overline{\underline{\left\|\ S\in\{1,2\}\ \right\|}} \) .