Fie \( z_{1},z_{2},...,z_{n}, n\geq 4 \), numere complexe(nu neaparat distincte), pentru care \( |z_{1}|+|z_{2}|+...+|z_{n}|=1 \).
Demonstrati ca se pot alege unele dintre acestea pentru care modulul sumei este strict mai mare decat \( \frac{1}{4} \)
Shl 2003
|z_{1}|+|z_{2}|+...+|z_{n}|=1
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Theodor Munteanu
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Fie \( {\rm z}_k = a_k + ib_k ,k = \overline {1,n} \)
Fie \( {\rm M = \{ }z_1 ,z_2 ,...,z_n \} ,M_1 = \{ z \in M|{{\rm Re}\nolimits} z \ge 0\} ,M_2 = \{ z \in M|{{\rm Re}\nolimits} z \le 0\} \)
\( M_3 = \{ z \in M|{{\rm Im}\nolimits} z \ge 0\} ,M_4 = \{ z \in M|{{\rm Im}\nolimits} z \le 0\} \);
\( |a + bi| \le |a| + |b|,\sum\limits_{i = 1}^n {|z_i |} = 1 \Rightarrow \sum\limits_{k \in M_1 } {|a_k | + } \)\(
\sum\limits_{k \in M_2 } {|a_k | + \sum\limits_{k \in M_3 } {|b_k | + \sum\limits_{k \in M_4 } {|b_k | \ge 1 \rightarrow \) cel putin unul din numere e \( \geq \frac{1}{4} \).Presupunem ca prima suma e \( \geq \frac{1}{4} \)
Atunci \( \sum\limits_{k \in M_1 } {|a_k | = |\sum\limits_{k \in M} {a_k | \Rightarrow ||\sum {a_k | \ge \frac{1}{4}.} } } \).
Fie \( {\rm M = \{ }z_1 ,z_2 ,...,z_n \} ,M_1 = \{ z \in M|{{\rm Re}\nolimits} z \ge 0\} ,M_2 = \{ z \in M|{{\rm Re}\nolimits} z \le 0\} \)
\( M_3 = \{ z \in M|{{\rm Im}\nolimits} z \ge 0\} ,M_4 = \{ z \in M|{{\rm Im}\nolimits} z \le 0\} \);
\( |a + bi| \le |a| + |b|,\sum\limits_{i = 1}^n {|z_i |} = 1 \Rightarrow \sum\limits_{k \in M_1 } {|a_k | + } \)\(
\sum\limits_{k \in M_2 } {|a_k | + \sum\limits_{k \in M_3 } {|b_k | + \sum\limits_{k \in M_4 } {|b_k | \ge 1 \rightarrow \) cel putin unul din numere e \( \geq \frac{1}{4} \).Presupunem ca prima suma e \( \geq \frac{1}{4} \)
Atunci \( \sum\limits_{k \in M_1 } {|a_k | = |\sum\limits_{k \in M} {a_k | \Rightarrow ||\sum {a_k | \ge \frac{1}{4}.} } } \).
La inceput a fost numarul. El este stapanul universului.