Μιγαδικοί 13

mathxl · #1

13.Εάν ${z_1},{z_2},{z_3} \in C$ ώστε $\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = 1$ τότε να δείξετε ότι :
$\displaystyle{{{\rm{(}}{{\rm{z}}_1}{{\rm{z}}_2}{\rm{ + }}{{\rm{z}}_2}{{\rm{z}}_3}{\rm{ + }}{{\rm{z}}_3}{{\rm{z}}_1}{\rm{ + }}{{\rm{z}}_1}{\rm{ + }}{{\rm{z}}_2}{\rm{ + }}{{\rm{z}}_3}{\rm{)}}^2}{\rm{ = }}{{\rm{z}}_1}{{\rm{z}}_2}{{\rm{z}}_3}{\left| {{{\rm{z}}_1}{{\rm{z}}_2}{\rm{ + }}{{\rm{z}}_2}{{\rm{z}}_3}{\rm{ + }}{{\rm{z}}_3}{{\rm{z}}_1}{\rm{ + }}{{\rm{z}}_1}{\rm{ + }}{{\rm{z}}_2}{\rm{ + }}{{\rm{z}}_3}} \right|^2}}$

(1988 Romanian Math Olympiad, School Competition, 10th grade)

Παρακαλώ όχι λακωνικές απαντήσεις.

Leo · #2

$\left| {{z}_{i}} \right|=1\Rightarrow {{\overline{z}}_{i}}=\frac{1}{{{z}_{i}}}$ ,

$\displaystyle{{{\left| {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right|}^{2}}=}$

$\displaystyle{\left( {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)\left( \overline{{{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}}} \right)=}$

$\displaystyle{\left( {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)\left( \frac{1}{{{z}_{1}}{{z}_{2}}}+\frac{1}{{{z}_{2}}{{z}_{3}}}+\frac{1}{{{z}_{3}}{{z}_{1}}}+\frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}} \right)}$

$\displaystyle{\therefore {{z}_{1}}{{z}_{2}}{{z}_{3}}{{\left| {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right|}^{2}}=\left( {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)\left( {{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}+{{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)}$

mathxl · #3

Howdy Leo!! Thank you very much!
Χαιρέτισα και ευχαρίστησα.

Μιγαδικοί 13

Μιγαδικοί 13

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