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Urmator
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andrei_bulaichi
user (0)
Pe: 2015-07-15T18:16:03+03:00 In: In:

Nr complexe 128

Se\ considera\ numerele\ complexe\ z_1,z_2,z_3\ care\ indeplinesc\ conditiile\\ |z_1|=|z_2|=|z_3|=1\ , \ z_1+ z_2+ z_3 \ne 0\ si\ z_1^{2}+ z_2^{2}+ z_3^{2}=0.\ Atunci\ |z_1+ z_2+ z_3|\ are \ valoarea:\\ a)1\ \ \ \ \ b)0\ \ \ \ \ c)4\ \ \ \ \ d)8\ \ \ \ \ e)2\ \ \ \ \ f)6

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3 raspunsuri

  1. expert (VI)

    Avem (ne putem gandi la ridicarea la patrat urmatoare ca o relatie ca una ce leaga atat expresia de calculat, cat si egalitatea din ipoteza) (z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2=2(z_1z_2+z_2z_3+z_3z_1)=2(z_1z_2+z_2z_3+z_3z_1), deci 

    *** QuickLaTeX cannot compile formula:
|z_1+z_2+z_3|^2=2|z_1z_2+z_2z_3+z_3z_1|=2|\overline{z_1z_2+z_2z_3+z_3z_1}|=2\left| \frac{1}{z_1z_2}+\frac{1}{z_2z_3}=\frac{1}{z_3z_1\right|=2\frac{|z_1+z_2+z_3|}{|z_1|\cdot |z_2|\cdot |z_3|}=2|z_1+z_2+z_3|

*** Error message:
File ended while scanning use of \frac .
Emergency stop.

    . Cum |z_1+z_2+z_3|\neq 0, impartind cu aceast numar obtinem |z_1+z_2+z_3|=2.

    NOTA: Am folosit faptul ca daca |z|=1 \Rightarrow \bar z = \frac{1}{z}.

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  2. user (0)

    Va multumesc ! 😀

    • 0
  4. expert (VI)

    Cu drag! 😀 Am editat postul anterior. Raspunsul corect este 2, m-am gandit rau cand am impartit.

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