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MirunaP
Pe: 2013-01-02T12:26:36+02:00 In: In:

Numere complexe in forma trigonometrica

Daca

    \[z + \frac{1}{z} = -2cosx\]

, sa se calculeze z^{n} + \frac{1}{z^{n}} .

Am incercat sa aduc prima ecuatie la forma unei ecuatii de gr II, dar delta mi-a iesit urat. Poate am gresit, poate nu e ideea de rezolvare buna, dar nu stiu cum sa fac mai departe.
Multumesc anticipat!

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3 raspunsuri

  1. George_Gaumont

    Ai o varianta de abordare.

    Attached files

    ecuatie_trigonometrica_162.pdf (62.9 KB) 

  2. gunty
    maestru (V)

    Ideea ta de rezolvare este buna, doar trebuie facuta pana la capat, de exemplu in felul urmator:

        \[ \begin{array}{l} z + \frac{1}{z} = - 2\cos x \\ \Leftrightarrow z^2 + 1 = - 2z \cdot \cos x \\ \Leftrightarrow z^2 + \left( {2\cos x} \right)z + 1 = 0 \\ \Delta = 4\cos ^2 x - 4 = 4\left( {\cos ^2 x - 1} \right) = - 4\sin ^2 x \\ \Rightarrow z_{1,2} = \frac{{ - 2\cos x \pm 2i \cdot \sin x}}{2} = - \cos x \pm i \cdot \sin x \\ \Rightarrow z_1 = - \left( {\cos x - i \cdot \sin x} \right) = - \left[ {\cos \left( { - x} \right) + i \cdot \sin \left( { - x} \right)} \right]_{\left( {*1} \right)} \\ \Rightarrow z_2 = - \left( {\cos x + i \cdot \sin x} \right)_{\left( {*2} \right)} \\ \left( {*1} \right),\left( {*2} \right) \Rightarrow z_{1,2} = - \left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right] \\ \end{array} \]

    Deci inlocuim

        \[ z = - \left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right] \]

    :

        \[ \begin{array}{l} \Rightarrow z^n + \frac{1}{{z^n }} = \left\{ { - \left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right]} \right\}^n + \frac{1}{{\left\{ { - \left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right]} \right\}^n }} = \\ = \left( { - 1} \right)^n \left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right]^n + \frac{1}{{\left( { - 1} \right)^n \left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right]^n }} = \\ = \left( { - 1} \right)^n \left\{ {\left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right]^n + \frac{1}{{\left[ {\cos \left( { \pm x} \right) + i \cdot \sin \left( { \pm x} \right)} \right]^n }}} \right\} \\ \end{array} \]

    Aplicam formula de Moivre, si primim:

        \[ \begin{array}{l} z^n + \frac{1}{{z^n }} = \left( { - 1} \right)^n \left[ {\cos \left( { \pm nx} \right) + i \cdot \sin \left( { \pm nx} \right) + \frac{1}{{\cos \left( { \pm nx} \right) + i \cdot \sin \left( { \pm nx} \right)}}} \right] = \\ = \left( { - 1} \right)^n \left\{ {\cos \left( { \pm nx} \right) + i \cdot \sin \left( { \pm nx} \right) + \frac{{\cos \left( { \pm nx} \right) - i \cdot \sin \left( { \pm nx} \right)}}{{\left[ {\cos \left( { \pm nx} \right) + i \cdot \sin \left( { \pm nx} \right)} \right]\left[ {\cos \left( { \pm nx} \right) - i \cdot \sin \left( { \pm nx} \right)} \right]}}} \right\} = \\ = \left( { - 1} \right)^n \left[ {\cos \left( { \pm nx} \right) + i \cdot \sin \left( { \pm nx} \right) + \frac{{\cos \left( { \pm nx} \right) - i \cdot \sin \left( { \pm nx} \right)}}{{\cos ^2 \left( { \pm nx} \right) + \sin ^2 \left( { \pm nx} \right)}}} \right] = \\ = \left( { - 1} \right)^n \left[ {\cos \left( { \pm nx} \right) + i \cdot \sin \left( { \pm nx} \right) + \cos \left( { \pm nx} \right) - i \cdot \sin \left( { \pm nx} \right)} \right] = \\ = \left( { - 1} \right)^n \cdot 2\cos \left( { \pm nx} \right) = \left( { - 1} \right)^n \cdot 2\cos \left( {nx} \right) \\ \end{array} \]

    In concluzie,

        \[ z^n + \frac{1}{{z^n }} = \left( { - 1} \right)^n \cdot 2\cos \left( {nx} \right) \]

    .

  4. MirunaP

    Am inteles, multumesc mult 🙂

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