101. O problema cu numere complexe.

by Virgil Nicula, Sep 7, 2010, 4:30 PM

Quote:
$\boxed{\ \begin{array}{c} \{z_1,z_2,z_3\}\subset\mathbb C\\\\ z=\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1\end{array}\ }\ \Longleftrightarrow\ \boxed {\ Re(z)+\sqrt{3}\cdot \left|Im(z)\right|\le |z_1|^2+|z_2|^2+|z_3|^2\ }$

Demonstratie. Se stie sau se demonstreaza usor ca aria $S$ a triunghiului $A_1A_2A_3$ , unde $A_k(z_k)$ , $k\in\overline {1,3}$ este data de relatia (retineti !) $\overline {\underline {\left\|\ S=\frac 14\cdot\left|z-\overline z\right|=\frac 12\cdot\left|Im(z)\right|\mathrm {\ ,\ unde\ }z=\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1\ \right\|}}\ \ (*)$ . Acum "traducem" in limbajul numerelor complexe cunoscuta inegalitate Weitzenbock $\underline {\overline {\left\|\ a^2+b^2+c^2 \ge 4\sqrt 3\cdot S\ \right\|}}$ (vezi http://en.wikipedia.org/wiki/Weitzenb%C3%B6ck%27s_inequality ) : $|z_1-z_2|^2+|z_2-z_3|^2+|z_3-z_1|^2\ge 2\sqrt 3\cdot |Im(z)|$ $\Longleftrightarrow$ $2\sum|z_k|^2-z-\overline z\ge 2\sqrt 3\cdot\left|Im(z)\right|$ $\Longleftrightarrow$ $2\sum|z_k|^2-2\cdot Re(z)\ge 2\sqrt 3\cdot\left|Im(z)\right|$ $\Longleftrightarrow$ $Re(z)+\sqrt{3}\cdot \left|Im(z)\right|\le |z_1|^2+|z_2|^2+|z_3|^2$ . Am folosit faptul ca $|z|^2=z\cdot\overline {z}$ si in particular $|z_1-z_2|^2=(z_1-z_2)(\overline {z_1}-\overline {z_2})=|z_1|^2+|z_2|^2-z_1\overline {z_2}-\overline {z_1}z_2$ etc.

Comentariu. S-au folosit exprimarea ariei unui triunghi cu ajutorul afixelor varfurilor acestuia si o inegalitate remarcabila. Problema propusa de fapt este o "traducere" si nu o "compozitie" pentru cel care cunoaste semnificatia geometrica a lui $z=\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1$ in aplicarea numerelor complexe in geometrie. De exemplu, punctele $A_k(z_k)$ , $k\in\overline {1,3}$ sunt coliniare daca si numai daca $Im(z)=0$ . Elegant exprimat (clasa a XI - a) , $A_3\in A_1A_2 \ \Longleftrightarrow\ \left|\begin{array}{ccc} z_1 & \overline {z_1} & 1\\\ z_2 & \overline {z_2} & 1\\\ z_3 & \overline {z_3} & 1\end{array}\right|=0$ .

Invitatie. Incercati sa compuneti probleme in mod asemanator : "traduceti" in limbajul numerelor complexe alte inegalitati geometrice cunoscute astfel incat "produsul" sa fie agreabil, adica sa arate frumos. De exemplu, inegalitatea Hadwiger-Finsler $\underline {\overline {\left\|\ a^2+b^2+c^2\ge4\sqrt 3\cdot S+(a-b)^2+(b-c)^2+(c-a)^2\ \right\|}}$ care este o extindere a inegalitatii Weitzenbock. Veti obtine o extindere a inegalitatii propuse. Poate are si aspect frumos. Incercati ! Este un exercitiu foarte bun pentru insusirea aplicarii numerelor complexe in geometrie.

Un exercitiu usor (clasa a XI - a). $z_k=x_k+i\cdot y_k\ ,\ k\in\overline {1,3}\ \Longrightarrow\ \mathrm {mod} \left|\begin{array}{ccc} x_1 & y_1 & 1\\\ x_2 & y_2 & 1\\\ x_3 & y_3 & 1\end{array}\right|=\frac 12\cdot\mathrm {mod}\left|\begin{array}{ccc} z_1 & \overline {z_1} & 1\\\ z_2 & \overline {z_2} & 1\\\ z_3 & \overline {z_3} & 1\end{array}\right|\ .$

Obtineti astfel demostratia formulei $(*)$ a ariei unui triunghi cu numere complexe (afixele varfurilor acestuia).
This post has been edited 1 time. Last edited by Virgil Nicula, Nov 23, 2015, 8:31 AM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404396
  • Total comments: 37
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