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MirunaP
Pe: 2013-01-03T15:21:44+02:00 In: In:

Ecuatii binome

Sa se rezolve in multimea C ecuatiile :

a) z^{4}=\frac{2-i}{2+\sqrt{3}+i(2\sqrt{3}-1)}

b) z^{6}-z^{5}+z^{4}-z^{3}+z^{2}-z+1=0

La prima am calculat, ca deobicei, sinus si cosinus de phi pentru z1(numaratorul) si z2(numitorul), dar mi-a dat niste valori care nu corespundeau niciunui unghi folosit deobicei ..
Multumesc anticipat!

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2 raspunsuri

  1. gunty
    maestru (V)

    a.

        \[ \begin{array}{l} \frac{{2 - i}}{{2 + \sqrt 3 + i\left( {2\sqrt 3 - 1} \right)}} = \frac{{\left( {2 - i} \right)\left( {2 + \sqrt 3 - i\left( {2\sqrt 3 - 1} \right)} \right)}}{{\left( {2 + \sqrt 3 } \right)^2 + \left( {2\sqrt 3 - 1} \right)^2 }} = \\ = \frac{{\left( {2 - i} \right)\left( {2 + \sqrt 3 - 2\sqrt 3 i + i} \right)}}{{20}} = \frac{{1 - \sqrt 3 i}}{4} = \frac{1}{4} + \frac{{ - \sqrt 3 }}{4}i \\ \left| {\frac{1}{4} + \frac{{ - \sqrt 3 }}{4}i} \right| = \sqrt {\frac{1}{{16}} + \frac{3}{{16}}} = \sqrt {\frac{1}{4}} = \frac{1}{2} \\ \Rightarrow \frac{1}{4} + \frac{{ - \sqrt 3 }}{4}i = \frac{1}{2}\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}} \right) = \frac{1}{2}\left( {\cos 300 + i \cdot \sin 300} \right) = \frac{1}{2}\left( {\cos \frac{{5\pi }}{3} + i \cdot \sin \frac{{5\pi }}{3}} \right) \\ \end{array} \]

    Acum ai numarul in forma trigonometrica. Din cate stiu, exista ceva formula, cu care o poti rezolva in continuare. Bafta !:D

  2. MirunaP

    Multumesc, am inteles 😀

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