mateforum.ro

Fie \( z_1,z_2,z_3\in\mathbb{C},\ |z_1|=|z_2|=|z_3|=1,\ z_1^3+z_2^3+z_3^3+z_1z_2z_3=0 \). Demonstrati ca \( |z_1+z_2+z_3|\in\{1,2\} \).

Daniel Jinga, OLM Arges

Mie din calcule imi iese doar ca \( |z_{1}+z_{2}+z_{3}|=2 \). Acum postez solutia mea si astept sa vad de ce nu-mi da si 1.

Conjugam relatia din enunt si prin aducere la acelasi numitor obtinem \( z_{1}^{3}z_{2}^{3}+z_{2}^{3}z_{3}^{3}+z_{3}^{3}z_{1}^{3}+z_{1}^{2}z_{2}^{2}z_{3}^{2}=0\Leftrightarrow z_{1}^{3}z_{2}^{3}+z_{2}^{3}z_{3}^{3}+z_{3}^{3}z_{1}^{3}=-z_{1}^{2}z_{2}^{2}z_{3}^{2}\ (1) \).
Din relatia din eununt obtinem \( z_{1}^{3}+z_{2}^{3}+z_{3}^{3}=-z_{1}z_{2}z_{3} \). Ridicam la patrat aceasta ultima relatie si folosind \( (1) \) obtinem ca \( z_{1}^{6}+z_{2}^{6}+z_{3}^{6}=3z_{1}^{2}z_{2}^{2}z_{3}^{2}\Leftrightarrow (z_{1}^{2}+z_{2}^{2}+z_{3}^{2})(z_{1}^{4}+z_{2}^{4}+z_{3}^{4}-z_{1}^{2}z_{2}^{2}-z_{2}^{2}z_{3}^{2}-z_{3}^{2}z_{1}^{2})=0 \)\( \Rightarrow z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=0 \) sau \( z_{1}^{4}+z_{2}^{4}+z_{3}^{4}-z_{1}^{2}z_{2}^{2}-z_{2}^{2}z_{3}^{2}-z_{3}^{2}z_{1}^{2}=0 \).

Cazul 1. Daca \( z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=0 \) \( \Rightarrow \) (fiind o problema cunoscuta) \( |z_{1}+z_{2}+z_{3}|=2 \).

Cazul 2. Daca \( z_{1}^{4}+z_{2}^{4}+z_{3}^{4}-z_{1}^{2}z_{2}^{2}-z_{2}^{2}z_{3}^{2}-z_{3}^{2}z_{1}^{2}=0 \), atunci \( z_{1}^{2},z_{2}^{2},z_{3}^{2} \) sunt afixele varfurilor unui triunghi echilateral cu centrul in originea sistemului de axe \( \Rightarrow z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=0 \), acest caz reducandu-se la cazul precedent \( \Rightarrow |z_{1}+z_{2}+z_{3}|=2 \)

INCEARCA SA AJUNGI LA UN PRODUS EGAL CU ZERO, DE TIPUL (Z1 LA CUB +Z2 LA CUB)(Z2 LA CUB+Z3 LA CUB)(Z3 LA CUB + Z1 LA CUB)=0

DACA AJUNGI AICI ANUNTA-MA!

Fie \( s=z_1+z_2+z_3,\ q=z_1z_2+z_2z_3+z_3z_1,\ p=z_1z_2z_3 \).
Din relatia initiala si conjugata ei obtinem :
\( s^3-3qs=-4p \)
\( \frac{q^3}{p}-3qs=-4p \)
de unde avem ca \( s^3=\frac{q^3}{p}\Rightarrow \frac{s^3}{p}=|s|^3 \) caci \( |s|^2=s\overline{s}=\frac{sq}{p} \) si acum daca impartim prima relatie la \( p \):
\( |s|^3-3|s|^2+4=0\Leftrightarrow (|s|+1)(|s|-2)^2=0 \) de unde \( |s|=2.
\)

Am gresit acolo cand am scos o radacina patratica:
\( \frac{s^3}{p}=\pm |s|^3 \) si rezolvand ecuatia \( -|s|^3-3|s|^2+4=0\Leftrightarrow (|s|-1)(|s|+2)^2=0\Rightarrow |s|=1. \)

andy crisan wrote:Cazul 2. Daca \( z_{1}^{4}+z_{2}^{4}+z_{3}^{4}-z_{1}^{2}z_{2}^{2}-z_{2}^{2}z_{3}^{2}-z_{3}^{2}z_{1}^{2}=0 \), atunci \( z_{1}^{2},z_{2}^{2},z_{3}^{2} \) sunt afixele varfurilor unui triunghi echilateral cu centrul in originea sistemului de axe

Asta e doar partial adevarat! Mai avem inca o posibilitate: \( z_1^2=z_2^2=z_3^2 \).

Aveti dreptate: analizand acest ultim caz obtinem ca doua dintre numere sunt opuse (se arata usor ca toate nu pot fi egale) \( \Rightarrow \) \( |z_{1}+z_{2}+z_{3}|=|z_{1}|=1 \).

EN18. \( \ \ \ \left\{\ z_1\ ,\ z_2\ ,\ z_3\ \right\}\ \subset\ \mathbb{C}\ \ \wedge\ \ \left\|\ \begin{array}{c}
|z_1|=|z_2|=|z_3|=1\\\\\\
z_1^3+z_2^3+z_3^3+z_1z_2z_3=0\end{array}\ \right\|\ \Longrightarrow\ |z_1+z_2+z_3|\ \in\ \{\ 1\ ,\ 2\ \}\ . \)

PR18. \( \ \ \ \frac {z_1}{u}=\frac {z_2}{v}=\frac {z_3}{1}=\frac {z_1+z_2+z_3}{u+v+1}\ \Longrightarrow\ \left\|\ \begin{array}{c}
|u|=|v|=1\\\\\\\\
u^3+v^3+1+uv=0\ (1)\\\\\\\\
\left|z_1+z_2+z_3\right|=|u+v+1|\end{array}\ \right\|\ . \)

\( (1)\ \ \wedge\ \ \overline u=\frac 1u\ \ \wedge\ \ \overline v=\frac 1v\ \Longrightarrow\ u^3+v^3+u^3v^3+u^2v^2=0\ (2)\ . \)

\( (1)\ \ \wedge\ \ (2)\ \Longrightarrow\ 1+uv=u^2v^2(1+uv)=-\left(u^3+v^3\right)\ \Longrightarrow \)

\( (1+uv)^2(1-uv)=0\ \ \wedge\ \ \left(u^3+v^3\right)+(1+uv)=0\ . \)

\( \odot\ \ uv=1\ \wedge\ \left(u^3+v^3\right)+2=0\ \Longrightarrow\ (u+v)^3-3(u+v)+2=0\ \Longrightarrow\ u+v=1\ \vee\ u+v=-2\ \Longrightarrow \)

\( u+v+1\in\{2,-1\}\ \Longrightarrow\ |u+v+1|\in \{1,2\}\ \Longrightarrow\ \left|z_1+z_2+z_3\right|\in\{1,2\}\ . \)

\( \odot\ \ uv=-1\ \wedge\ u^3+v^3=0\ \Longrightarrow\ (u+v)^3+3(u+v)=0\ \Longrightarrow\ u+v\in\left\{0,\pm i\sqrt 3\right\}\ \Longrightarrow \)

\( u+v+1\in \left\{1,1\pm i\sqrt 3\right\}\ \Longrightarrow\ |u+v+1|\in\{1,2\}\ \Longrightarrow\ \left|z_1+z_2+z_3\right|\in\{1,2\}\ . \)

\( \left\|\ \begin{array}{c}
|u|=|v|=1\\\\\\
u^3+v^3+1+uv=0\end{array}\ \right\|\ \Longrightarrow\ u+v\ \in\ \left\{\ -2\ ,\ 1\ ,\ 0\ ,\ \pm i\sqrt 3\ \right\}\ . \)

\( \left\{\ \begin{array}{c}
u+v=1\\\\\\
uv=1\end{array}\ \ \vee\ \ \left\{\ \begin{array}{c}
u+v=-2\\\\\\
uv=1\end{array}\ \ \vee\ \ \left\{\ \begin{array}{c}
u+v=0\\\\\\
uv=-1\end{array}\ \ \vee\ \ \left\{\ \begin{array}{c}
u+v=i\sqrt 3\\\\\\
uv=-1\end{array}\ \ \vee\ \ \left\{\ \begin{array}{c}
u+v=-i\sqrt 3\\\\\\
uv=-1\end{array}\ \ . \)