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grapefruit
maestru (V)
Pe: 2013-09-10T15:33:44+03:00 In: In:

Numar de solutii in C

Ecuatia z^2=z conjugat are in multimea C un numar de …. solutii.

Am incercat mai multe variante insa nu am reusit.
Am zis ca fie z=a+bi,z bara=a-bi

a^2 + 2abi-b^2=a-bi
a^2 -b^2 + 2abi=a-bi
Rezulta sistemul:
a^2-b^2=a(1)
2ab=-b(2)
(2) 2ab +b=0
b(2a+1)=0 b=0;a=-1/2
…bla bla bla si nu stiu cum sa continui.

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  1. gunty
    maestru (V)

    Ai inceput foarte bine, doar trebuie facut pana la capat:

        \[ \begin{array}{l} z^2 = \overline z _{\left( 1 \right)} . \\ Fie\;z = a + bi. \Rightarrow \left( 1 \right) \Leftrightarrow a^2 - b^2 + 2abi = a - bi \Leftrightarrow a^2 - a - b^2 + i\left( {2ab + b} \right) = 0. \Rightarrow \left\{ \begin{array}{l} a^2 - a - b^2 = 0 \\ 2ab + b = 0 \\ \end{array} \right. \\ Daca\;b = 0 \Rightarrow a^2 - a = 0 \Leftrightarrow a\left( {a - 1} \right) = 0 \Rightarrow a = 0 \vee a = 1. \\ Daca\;b \ne 0 \Rightarrow \left\{ \begin{array}{l} a^2 - a - b^2 = 0 \\ 2a + 1 = 0 \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a^2 - a - b^2 = 0 \\ a = - \frac{1}{2} \\ \end{array} \right.. \Rightarrow \frac{1}{4} + \frac{1}{2} - b^2 = 0 \Leftrightarrow b^2 - \frac{3}{4} = 0 \Rightarrow b = \pm \frac{{\sqrt 3 }}{2} \\ Deci\;solutiile\;ecuatiei\;\left( 1 \right)\;sunt\;\left\{ \begin{array}{l} z_1 = 0 + 0i = 0 \\ z_2 = 1 + 0i = 1 \\ z_3 = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i \\ z_4 = - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i \\ \end{array} \right.,\;adica\;4\;la\;numar. \\ \end{array} \]

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