Răspuns:
Explicație pas cu pas:
Asta e o problema de aia clasica .
Toata schema era sa observi ca:
[tex](z_1+z_2)(z_2+z_3)(z_3+z_1)+z_1z_2z_3=(z_1+z_2+z_3)(z_1z_2+z_2z_3+z_3z_1)\\\text{Din egaltitatea } (z_1+z_2+z_3)(z_1z_2+z_2z_3+z_3z_1)=0\text{ se obtin doua}\\\text{posibilitati:}\\1. z_1+z_2+z_3=0\\\text{Ei bine asta e o proprietate destul de cunoscuta. Se demonstreaza f. usor }\\\text{Fie }|z_1|=|z_2|=|z_3|=r, r>0.\\z_1+z_2+z_3=0\\z_1+z_2=-z_3\\|z_1+z_2|=|z_3|\\|z_1+z_2|^2=|z_3|^2\\(z_1+z_2)(\overline{z_1}+\overline{z_2})=r^2[/tex]
[tex]z_1\overline{z_1}+z_1\overline{z_2}+z_2\overline{z_1}+z_2\overline{z_2}=r^2\\|z_1|^2+|z_2|^2+z_1\overline{z_2}+z_2\overline{z_1}=r^2\\2r^2+z_1\overline{z_2}+z_2\overline{z_1}=r^2\\z_1\overline{z_2}+z_2\overline{z_1}=-r^2 .\\\text{Pe de alta parte avem ca: }\\|z_1-z_2|^2=(z_1-z_2)(\overline{z_1}-\overline{z_2})=|z_1|^2+|z_2|^2-z_1\overline{z_2}-z_2\overline{z_1}=2r^2+r^2=3r^2\\\text{Deci }|z_1-z_2|=r\sqrt 3.\\\text{Analog se demonstreaza ca }|z_2-z_3|=|z_3-z_1|=r\sqrt 3,\text{deci triunghiul}[/tex]
[tex]\text{ este echilateral.}\\\text{Cea dea doua posibilitatea este }\\z_1z_2+z_2z_3+z_3z_1=0\\\text{Din nou vom nota cu r modulul.}\\|z_1|=r\Rightarrow \overline{z_1}=\dfrac{r}{z_1}\\\text{Putem conjuga toata relatia:}\\\overline{z_1z_2+z_2z_3+z_3z_1}=0\\\overline{z_1}\cdot\overline{z_2}+\overline{z_2}\cdot\overline{z_3}+\overline{z_3}\cdot\overline{z_1}=0\\\dfrac{r}{z_1}\cdot\dfrac{r}{z_2}+\dfrac{r}{z_2}\cdot\dfrac{r}{z_3}+\dfrac{r}{z_3}\cdot\dfrac{r}{z_1}=0[/tex]
[tex]r^2\left(\dfrac{1}{z_1z_2}+\dfrac{1}{z_2z_3}+\dfrac{1}{z_3z_1}\right)=0\\\dfrac{1}{z_1z_2}+\dfrac{1}{z_2z_3}+\dfrac{1}{z_3z_1}=0 |\cdot z_1z_2z_3\\z_3+z_2+z_1=0\\\text{Si ajungem in primul caz demonstrat anterior.Daca mai ai intrebari nu }\\\text{ezita sa intrebi.}[/tex]
