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Ronela
Pe: 2012-10-06T18:43:52+03:00 In: In:

Modulul unui numar complex

Am nevoie de ajutor sau macar o indicatie cat de repede…multumesc anticipat 🙂

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  1. PhantomR
    expert (VI)
  2. Integrator
    maestru (V)

    PhantomR wrote: |z_k|=r \Rightarrow z_k \cdot \bar z_k=|z_k^2=r^2. Cum |z_k|>0 \Rightarrow z_k\neq 0 si deci împărtind prin z_k relatia z_k \cdot \bar z_k = r^2 \Rightarrow \bar z_k=\frac{r^2}{z_k}.

    Mai departe, \bar E = \frac{(\bar z_1+\bar z_2)\cdot ... \cdot (\bar z_n+\bar z_1)}{\bar z_1\cdot ... \cdot \bar z_n}=\frac{\left( \frac{r^2}{z_1}+ \frac{r^2}{z_2} \right) \cdot ... \left ( \frac{r^2}{z_n}+\frac{r^2}{z_1}\right) }{\frac{r^2}{z_1}\cdot ... \cdot \frac{r^2}{z_n}}=\frac{\left[r^2\left(\frac{1}{z_1}+\frac{1}{z_2}\right)\right] \cdot ... \cdot \left [ r^2 \left( \frac{1}{z_n}+\frac{1}{z_1} \right) \right] }{\frac{(r^2)^n}{z_1...z_n}}=\frac{(r^2)^n\cdot \frac{z_2+z_1}{z_1z_2}\cdot ... \cdot \frac{z_1+z_n}{z_1z_n}}{\frac{r^{2n}}{z_1...z_n}}=\frac{r^{2n} \cdot \frac{(z_1+z_2)...(z_n+z_1)}{z_1z_2\cdot z_2z_3 \cdot ... z_nz_1}}{\frac{r^{2n}}{z_1z_2...z_n}}=\frac{\frac{(z_1+z_2)\cdot ... \cdot (z_n+z_1)}{(z_1z_2...z_n)^2}}{\frac{1}{z_1z_2...z_n}} =(z_1z_2...z_n)\cdot \frac{(z_1+z_2)...(z_n+z_1)}{(z_1z_2...z_n)^2}=\frac{(z_1+z_2)...(z_n+z_1)}{z_1...z_n}. Asadar E=\bar E \Rightarrow E\in\mathbb{R}.


    Fara suparare dar nu inteleg.Daca consideram ca n=2 si deciz_1=a+b \cdot i si z_2=c+d \cdot i astfel incat r=\sqrt {a^2+b^2}=\sqrt{c^2+d^2} atunci care e valoarea lui E?Multumesc!

  4. Integrator
    maestru (V)

    Ronela wrote: Am nevoie de ajutor sau macar o indicatie cat de repede…multumesc anticipat 🙂


    Fara suparare,dar nu inteleg.Sigur este asa enuntul problemei?Multumesc!

  5. Ronela

    Da, asa este enuntul:|

  6. Ronela

    Din acest motiv am fost si eu nedumerita:|

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