Fie \( z_1,z_2,z_3 \in \mathbb C \) astfel încât \( z_1+z_2+z_3=0. \)
a) Să se arate că dacă \( |z_1|=|z_2|=|z_3| \), atunci \( z_1z_2+z_2z_3+z_3z_1=0. \)
b) Să se arate că dacă \( z_1z_2+z_2z_3+z_3z_1=0, \) atunci \( |z_1|+ \varepsilon |z_2|+ \varepsilon^2 |z_3|=0, \) unde \( \varepsilon \in {\mathbb C} \setminus {\mathbb R}, \ \varepsilon^3=1. \)
Marius Perianu, OLM 2008 Olt
Problema simpla cu numere complexe
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
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Marius Perianu
- Euclid
- Posts: 40
- Joined: Thu Dec 06, 2007 11:40 pm
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Problema simpla cu numere complexe
Last edited by Marius Perianu on Sun Aug 23, 2009 1:59 pm, edited 2 times in total.
Daca \( z_{1,3}=0 \) problema este evidenta. Acum luam \( z_{1,3}\in \mathbb{C}^* \).
a) Notam \( |z_1|=|z_2|=|z_3|=r> 0 \).
Din \( z_1+z_2+z_3=0 \) rezulta \( \bar {z_1+z_2+z_3}=0 \).
Deci \( r^2(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})=r^2 \frac{z_1z_2+z_2z_3+z_3z_1}{z_1z_2z_3}=0 \) si cum \( r>0 \) atunci concludem \( z_1z_2+z_2z_3+z_3z_1=0 \).
b) La acest punct cred ca e scris gresit modul, altfel problema ar fi banala:
\( |z_1|+\varepsilon |z_2|+\varepsilon^2 |z_3|=|z_1|(1+\varepsilon+\varepsilon^2 )=0 \)
Voi demonstra relatia fara sa aplic modul la z-uri
.
Deoarece \( z_1z_2+z_2z_3+z_3z_1=z_1+z_2+z_3=0 \) atunci din \( (z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1) \) avem \( z_1^2+z_2^2+z_3^2=0 \).
Folosim identitatea:
\( z_1^2+z_2^2+z_3^2-(z_1z_2+z_2z_3+z_3z_1)=(z_1+\varepsilon z_2+\varepsilon^2 z_3)(z_1+\varepsilon^2 z_2+\varepsilon z_3) \).
Dar \( z_1^2+z_2^2+z_3^2-(z_1z_2+z_2z_3+z_3z_1)=0 \).
Asadar \( z_1+\varepsilon z_2+\varepsilon^2 z_3=0 \).
a) Notam \( |z_1|=|z_2|=|z_3|=r> 0 \).
Din \( z_1+z_2+z_3=0 \) rezulta \( \bar {z_1+z_2+z_3}=0 \).
Deci \( r^2(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3})=r^2 \frac{z_1z_2+z_2z_3+z_3z_1}{z_1z_2z_3}=0 \) si cum \( r>0 \) atunci concludem \( z_1z_2+z_2z_3+z_3z_1=0 \).
b) La acest punct cred ca e scris gresit modul, altfel problema ar fi banala:
\( |z_1|+\varepsilon |z_2|+\varepsilon^2 |z_3|=|z_1|(1+\varepsilon+\varepsilon^2 )=0 \)
Voi demonstra relatia fara sa aplic modul la z-uri
.Deoarece \( z_1z_2+z_2z_3+z_3z_1=z_1+z_2+z_3=0 \) atunci din \( (z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1) \) avem \( z_1^2+z_2^2+z_3^2=0 \).
Folosim identitatea:
\( z_1^2+z_2^2+z_3^2-(z_1z_2+z_2z_3+z_3z_1)=(z_1+\varepsilon z_2+\varepsilon^2 z_3)(z_1+\varepsilon^2 z_2+\varepsilon z_3) \).
Dar \( z_1^2+z_2^2+z_3^2-(z_1z_2+z_2z_3+z_3z_1)=0 \).
Asadar \( z_1+\varepsilon z_2+\varepsilon^2 z_3=0 \).
Last edited by Wizzy on Mon Mar 10, 2008 5:37 pm, edited 1 time in total.
Vrajitoarea Andrei
Re: Problema simpla cu numere complexe
b) Cam acelasi ideea (cu Wizzy) ( Btw .. interesanta identitatea ta ) .
Avem ca \( z_1,z_2,z_3 \) sunt radacinile polinomului \( f \in \mathbb{C}[X] \), unde \( f = Z^3 - (z_1+z_2+z_3)Z^2 + (z_1z_2 + z_2z_3 + z_3z_1)Z - z_1z_2z_3 \).
Adica \( f= Z^3 - z_1z_2z_3 \).
Luam pe rand \( f(z_1), f(z_2), f(z_3) \) si trecem la modul. Obtinem \( |z_1|=|z_2|=|z_3|=r \Rightarrow |z_1|+ \varepsilon|z_2|+ \varepsilon^2 |z_3|=r(1+\varepsilon + \varepsilon^2) = 0 \).
Geometric: Avem \( |z_1|=|z_2|=|z_3|=r \) si \( z_1 + z_2 + z_3 = 0 \). Atunci in triunghiul cu afixele respective avem \( O \equiv H \), adica triunghiul este echilateral \( \Longleftrightarrow z_1 + z_2\varepsilon + z_3\varepsilon^2 = 0 \).
) .Avem ca \( z_1,z_2,z_3 \) sunt radacinile polinomului \( f \in \mathbb{C}[X] \), unde \( f = Z^3 - (z_1+z_2+z_3)Z^2 + (z_1z_2 + z_2z_3 + z_3z_1)Z - z_1z_2z_3 \).
Adica \( f= Z^3 - z_1z_2z_3 \).
Luam pe rand \( f(z_1), f(z_2), f(z_3) \) si trecem la modul. Obtinem \( |z_1|=|z_2|=|z_3|=r \Rightarrow |z_1|+ \varepsilon|z_2|+ \varepsilon^2 |z_3|=r(1+\varepsilon + \varepsilon^2) = 0 \).
Geometric: Avem \( |z_1|=|z_2|=|z_3|=r \) si \( z_1 + z_2 + z_3 = 0 \). Atunci in triunghiul cu afixele respective avem \( O \equiv H \), adica triunghiul este echilateral \( \Longleftrightarrow z_1 + z_2\varepsilon + z_3\varepsilon^2 = 0 \).
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Marius Perianu
- Euclid
- Posts: 40
- Joined: Thu Dec 06, 2007 11:40 pm
- Location: Slatina
Marius Perianu wrote: b) Să se arate că dacă \( z_1z_2+z_2z_3+z_3z_1=0, \) atunci \( |z_1|+ \varepsilon |z_2|+ \varepsilon^2 |z_3|=0. \)
Numerele \( z_1=1, \ z_2=\varepsilon^2, \ z_3=\varepsilon \) iti arata de ce nu e scris gresit cu modul.Wizzy wrote:b) La acest punct cred ca e scris gresit modul ... Asadar \( z_1+\varepsilon z_2+\varepsilon^2 z_3=0. \)
Marius Perianu