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Laur
Pe: 2012-06-11T16:56:11+03:00 In: In:

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Sa se arate ca daca z1 si z2 apartin C(numere complexe), |z1|=|z2|=1 si |z1+z2|= radical din 3, atunci |z1-z2|=1.

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4 raspunsuri

  1. PhantomR
    expert (VI)

    Probabil ca intrebarea nu prea are legatura cu problema in sine, dar de ce postati o problema cu numere complexe la clasa a IX-a?

  2. PhantomR
    expert (VI)

    Este un desen mai jos. Vom interpreta vectorial numerele complexe. Scopul nostru este de a afla |z_1-z_2|=|\v{OD}|=OD.

    Notam \alpha = \angle{\left(\v{OA},\v{OB}\right)}.

    Uitandu-ne la paralelogramul AOBC avem OC^2=OA^2+OB^2+2OA\cdot OB\cos \alpha.

    Uitandu-ne la paralelogramul AOB'D avem, deoarece |\v{OB'}|=|-\v{OB}|=|\v{OB}|=OB, | OD^2=OA^2+OB^2-2OA \cdot OB \cos (\pi - \alpha)=OA^2+OB^2-2OA \cdot OB \cdot (-\cos \alpha)=OA^2+OB^2+2OA \cdot OB \cos \alpha.

    Adunand cele doua relatii rezulta OC^2+OD^2=2(OA^2+OB^2) \Rightarrow OD^2=2(OA^2+OB^2)-OC^2 \Rightarrow OD=\sqrt{2(OA^2+OB^2)-OC^2}=\sqrt{2(1^2+1^2)-\sqrt{3}^2}=1.

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  4. PhantomR
    expert (VI)

    Hmmm… Demonstratia 2:
    Vom folosi relatia z\bar z =|z|^2.

    Avem 1=1^2=|z_1+z_2|^2=(z_1+z_2)(\bar z_1+\bar z_2)=z_1\bar z_1+z_1\bar z_2+z_2\bar z_1+z_2\bar z_2 \Rightarrow z_1\bar z_2+z_2\bar z_1=1-z_1\bar z_1-z_2\bar z_2=1-1^2-1^2=-1.

    Acum |z_1-z_2|^2=(z_1-z_2)(\bar z_1-\bar z_2)=z_1\bar z_2 - z_1\bar z_2-z_2\bar z_1+z_2\bar z_2=|z_1|^2-(z_1\bar z_2+z_2\bar z_1)+|z_2|^2=1-1+1=1.

  5. Integrator
    maestru (V)

    Laur wrote: Sa se arate ca daca z1 si z2 apartin C(numere complexe), |z1|=|z2|=1 si |z1+z2|= radical din 3, atunci |z1-z2|=1.


    Fie z_1=a+bi si z_2=c+di si deci rezulta ca a^2+b^2=c^2+d^2=1 si a \cdot b+c \cdot d=\frac{1}{2} ceea ce inseamna ca |z_1-z_2|=sqrt{(a-b)^2+(c-d)^2}=sqrt{2-2 \cdot \frac{1}{2}}=1.

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