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równanie zespolone


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#1 kinia888

kinia888

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Napisano 15.03.2010 - 21:58

Zalozmy ze liczba \zeta spelnia rownanie  1+\zeta + \zeta^{2}=0
Wykazac ze dla dowolnych liczb zespolonych \alpha,\beta,\gamma ma miejsce rownosc
 3(|\alpha|^{2}+|\beta|^{2}+|\gamma|^{2})=|a|^{2}+|b|^{2}+|c|^{2}
gdzie  a=\alpha + \beta + \gamma,   b=\alpha + \zeta\beta +\zeta^{2}\gamma,   c=\alpha+ \zeta^{2}\beta + \zeta \gamma$
Prosze o jakies wskazówki
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Afroman

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Napisano 25.09.2011 - 17:55

#2 Kinia7

Kinia7

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Napisano 01.03.2017 - 00:00

\zeta=-\fr12\pm\fr{\sq3}{2}i\ \ \ \ \ \zeta^2=-\fr{1}{2}\mp\fr{\sq3}{2}i \quad\to\quad\ \{\zeta^2=\overline{\zeta}\\|\zeta|=1\\|\zeta^2|=1
           \fbox{\ |z|^2=z\cd\overline{z}\ }                        \fbox{\ \ \ \ \ \ \ \ \ \ \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\ \ \ \ \ \ \ \ \ \ \ \overline{z_1z_2}=\overline{z_1}\cd\overline{z_2}\\\ z_1\overline{z_2}+\overline{z_1}z_2=2\[Re(z_1)Re(z_2)+Im(z_1)Im(z_2)\]\ }
|a|^2=\(\alpha+\beta+\gamma\)\(\overline\alpha+\overline\beta+\overline\gamma\)=\alpha\overline\alpha+\alpha\overline\beta+\alpha\overline\gamma+\beta\overline\alpha+\beta\overline\beta+\beta\overline\gamma+\gamma\overline\alpha+\gamma\overline\beta+\gamma\overline\gamma=
\ \ \ \ \ \ =|\alpha|^2+\alpha\overline\beta+\alpha\overline\gamma+\beta\overline\alpha+|\beta|^2+\beta\overline\gamma+\gamma\overline\alpha+\gamma\overline\beta+|\gamma\|^2=
\ \ \ \ \ \ =|\alpha|^2+|\beta|^2+|\gamma\|^2+\alpha\overline\beta+\overline\alpha\beta+\alpha\overline\gamma+\overline\alpha\gamma+\beta\overline\gamma+\overline\beta\gamma
|b|^2=\(\alpha+\zeta\beta+\zeta^2\gamma\)\(\overline\alpha+\overline\zeta\overline\beta+\overline{\zeta^2}\overline\gamma\)=\(\alpha+\zeta\beta+\overline\zeta\gamma\)\(\overline\alpha+\overline\zeta\overline\beta+\zeta\overline\gamma\)=
\ \ \ \ \ \ =\alpha\overline\alpha+\alpha\overline\zeta\overline\beta+\alpha\zeta\overline\gamma+\overline\alpha\zeta\beta+\zeta\overline\zeta\beta\overline\beta+\zeta^2\beta\overline\gamma+\overline\alpha\overline\zeta\gamma+\overline\zeta^2\overline\beta\gamma+\zeta\overline\zeta\gamma\overline\gamma=
\ \ \ \ \ \ =|\alpha|^2+\alpha\overline\zeta\overline\beta+\alpha\zeta\overline\gamma+\overline\alpha\zeta\beta+|\zeta|^2|\beta|^2+\overline\zeta\beta\overline\gamma+\overline\alpha\overline\zeta\gamma+\zeta\overline\beta\gamma+|\zeta|^2|\gamma|^2=
\ \ \ \ \ \ =|\alpha|^2+|\beta|^2+|\gamma\|^2+\alpha\overline\zeta\overline\beta+\overline\alpha\zeta\beta+\alpha\zeta\overline\gamma+\overline\alpha\overline\zeta\gamma+\overline\zeta\beta\overline\gamma+\zeta\overline\beta\gamma=
|c|^2=\(\alpha+\zeta^2\beta+\zeta\gamma\)\(\overline\alpha+\overline{\zeta^2}\overline\beta+\overline{\zeta}\overline\gamma\)=\(\alpha+\overline\zeta\beta+\zeta\gamma\)\(\overline\alpha+\zeta\overline\beta+\overline\zeta\overline\gamma\)=
\ \ \ \ \ \ =\alpha\overline\alpha+\alpha\zeta\overline\beta+\alpha\overline\zeta\overline\gamma+\overline\alpha\overline\zeta\beta+\overline\zeta\zeta\beta\overline\beta+\overline\zeta^2\beta\overline\gamma+\overline\alpha\zeta\gamma+\zeta^2\overline\beta\gamma+\zeta\overline\zeta\gamma\overline\gamma=
\ \ \ \ \ \ =|\alpha|^2+\alpha\zeta\overline\beta+\alpha\overline\zeta\overline\gamma+\overline\alpha\overline\zeta\beta+|\zeta|^2|\beta|^2+\zeta\beta\overline\gamma+\overline\alpha\zeta\gamma+\overline\zeta\overline\beta\gamma+|\zeta|^2|\gamma|^2=
\ \ \ \ \ \ =|\alpha|^2+|\beta|^2+|\gamma\|^2+\alpha\zeta\overline\beta+\overline\alpha\overline\zeta\beta+\alpha\overline\zeta\overline\gamma+\overline\alpha\zeta\gamma+\zeta\beta\overline\gamma+\overline\zeta\overline\beta\gamma
|a|^2+|b|^2+|c|^2=3\(|\alpha|^2+|\beta|^2+|\gamma|^2\)+(\alpha\overline\beta+\alpha\overline\zeta\overline\beta+\alpha\zeta\overline\beta)+(\overline\alpha\beta+\overline\alpha\zeta\beta+\overline\alpha\overline\zeta\beta)+
\ \ \ \ \ \ \ \ +(\alpha\overline\gamma+\alpha\zeta\overline\gamma+\alpha\overline\zeta\overline\gamma)+(\overline\alpha\gamma+\overline\alpha\overline\zeta\gamma+\overline\alpha\zeta\gamma)+(\beta\overline\gamma+\overline\zeta\beta\overline\gamma+\zeta\beta\overline\gamma)+(\overline\beta\gamma+\zeta\overline\beta\gamma+\overline\zeta\overline\beta\gamma)=
\ \ \ \ \ \ =3\(|\alpha|^2+|\beta|^2+|\gamma|^2\)+\alpha\overline\beta(1+\overline\zeta+\zeta)+\overline\alpha\beta(1+\zeta+\overline\zeta)+
\ \ \ \ \ \ \ \ +\alpha\overline\gamma(1+\zeta+\overline\zeta)+\overline\alpha\gamma(1+\overline\zeta+\zeta)+\beta\overline\gamma(1+\overline\zeta+\zeta)+\overline\beta\gamma(1+\zeta+\overline\zeta)=\[\ \\1+\zeta+\overline\zeta=0\\\ \]=
\ \ \ \ \ \ =3\(|\alpha|^2+|\beta|^2+|\gamma|^2\)

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