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raluca.w
Pe: 2012-02-14T14:30:47+02:00 In: In:

Numere complexe – Inductie matematica

Fie

    \[ z \in C \]

astfel incat |z|=1. Arata ca

    \[ n\left| {1 + z} \right| + \left| {1 + z^2 } \right| + \left| {1 + z^3 } \right| + ... + \left| {1 + z^{2n} } \right| + \left| {1 + z^{2n + 1} } \right| \ge 2n \]

, unde

    \[ n \in {\rm N} \]

.
Problema provine din manualul editurii Sigma – 30/146. Multumesc!

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1 raspuns

  1. andu_flavius95
    maestru (V)

    O solutie fara inductie matematica:

    Demonstratie: \\ \\ n\left| {1 + z} \right| + \left| {1 + z^2 } \right| + \left| {1 + z^3 } \right| + ... + \left| {1 + z^{2n} } \right| + \left| {1 + z^{2n + 1} } \right| = n\left| {1 + z} \right| + \left( {\left| {1 + z^2 } \right| + \left| {1 + z^3 } \right|} \right) + \left( {\left| {1 + z^4 } \right| + \left| {1 + z^5 } \right|} \right)... + \left( {\left| {1 + z^{2n} } \right| + \left| {1 + z^{2n + 1} } \right|} \right) \ge \\ \ge n\left| {1 + z} \right| + \left| {1 + z^2 - 1 - z^3 } \right| + \left| {1 + z^4 - 1 - z^5 } \right| + ... + \left| {1 + z^{2n} - 1 - z^{2n + 1} } \right| = n\left| {1 + z} \right| + \left| {z^2 } \right| \cdot \left| {1 - z} \right| + \left| {z^4 } \right| \cdot \left| {z - 1} \right| + ... + \left| {z^{2n} } \right| \cdot \left| {1 - z} \right| = \\ = n\left| {1 + z} \right| + n\left| {1 - z} \right| = n\left( {\left| {1 + z} \right| + \left| {1 - z} \right|} \right) \ge n\left( {\left| {1 + z - 1 + z} \right|} \right) = n \cdot \left| {2z} \right| = 2n\quad (q.e.d) \\ \end{array} \]

    Nota: Am folosit inegalitatile de la modul. 😀
    [/tex]

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