Trei numere complexe cu modulul sumei egal cu 1 sau 2

[mihai++]

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

[andy crisan]

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 \( \Rightarrow |z_{1}+z_{2}+z_{3}|=2 \)

[Sorin Ulmeanu]

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!

[mihai++]

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 \( -|s|^3-3|s|^2+4=0\Leftrightarrow (|s|-1)(|s|+2)^2=0\Rightarrow |s|=1. \)

[bae]

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: \( z_1^2=z_2^2=z_3^2 \).

[andy crisan]

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

[Virgil Nicula]

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

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

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

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

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

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

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

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

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

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

[Virgil Nicula]

OFF-topic. In general in viata este mai usor sa gresesti decat sa intelegi unde si de ce ai gresit.
Si in matematica se poate intampla sa gresesti, insa astfel apare o noua problema
- descoperirea greselii - care te proiecteaza spre alte metode sau chiar extinderea teoriei.
Cu alte cuvinte, greseala in matematica uneori este benefica. Insa aici in noua problema
aparuta ai "dezavantajul" ca trebuie sa te descurci singur, cum de altfel s-a si intamplat.