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mihaellafly
Pe: 2013-01-13T14:03:47+02:00 In: In:

forma trigonometrica a numerelor complexe

Sa se scrie sub forma trigonometrica numerele complexe:
1) -cosa-isina, a\in [0,\frac{\Pi }{2}]
2) sina-icosa, a\in [\frac{\Pi }{2},\Pi ]
Va rog frumos ajutati-ma ! Multumesc!

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  1. gunty
    maestru (V)

    Se stie ca forma trigonometrica a unui numar complex z arata in felul urmator:

        \[ z = \left| z \right| \cdot \left( {\cos t + i \cdot \sin t} \right) \]

    , unde

        \[ \left| z \right| \ge 0 \]

    si

        \[ t \in \left[ {0;2\pi } \right) \]

    1)

        \[ - \cos a - i \cdot \sin a = 1 \cdot \left[ {\cos \left( {\pi + a} \right) + i \cdot \sin \left( {\pi + a} \right)} \right] \]

    Deoarece 1>=0 si din

        \[ a \in \left[ {0;\frac{\pi }{2}} \right] \Rightarrow \pi + a \in \left[ {0;2\pi } \right) \]

    , inseamna ca numarul e in forma trigonometrica.

    2)

        \[ \sin a - i \cdot \cos a = 1 \cdot \left[ {\cos \left( {\frac{\pi }{2} - a} \right) - i \cdot \sin \left( {\frac{\pi }{2} - a} \right)} \right] = 1 \cdot \left[ {\cos \left( {a - \frac{\pi }{2}} \right) + i \cdot \sin \left( {a - \frac{\pi }{2}} \right)} \right] \]

    Deoarece 1>=0 si din

        \[ a \in \left[ {\frac{\pi }{2};\pi } \right] \Rightarrow a - \frac{\pi }{2} \in \left[ {0;2\pi } \right) \]

    , inseamna ca numarul e in forma trigonometrica.

    * Am folosit formulele:

        \[ \begin{array}{l} \cos x = - \cos \left( {\pi + x} \right) \\ \sin x = - \sin \left( {\pi + x} \right) \\ \sin x = \cos \left( {\frac{\pi }{2} - x} \right) \\ \cos x = \sin \left( {\frac{\pi }{2} - x} \right) \\ \cos \left( { - x} \right) = \cos x \\ \sin \left( { - x} \right) = - \sin x \\ \end{array} \]

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