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#1 Kinia7

Kinia7

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Napisano 07.03.2016 - 09:23

Wykaż tożsamość     3(|\alpha|^2+|\beta|^2+|\gamma|^2)=|a|^2+|b|^2+|c|^2
jeżeli     a=\alpha+\beta+\gamma,\ \ b=\alpha+\zeta\beta+\zeta^2\gamma,\ \ c=\alpha+\zeta^2\beta+\zeta\gamma
a  \zeta  jest pierwiastkiem równania  \zeta^2+\zeta+1=0
 

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Napisano 25.09.2011 - 17:55

#2 bb314

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Napisano 10.03.2016 - 19:54

*
Najwyższa ocena

\bl 1+\zeta+\zeta^2=0                                                                                                          \bl\fbox{\ \ \ \ \ \ \ |z|^2=z\cd\overline{z}\\\ \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\ \\ \ \ \ \ \ \overline{z_1z_2}=\overline{z_1}\cd\overline{z_2}\ }
 
\zeta=-\fr12\pm\fr{\sq3}{2}i\gr\ \Rightarrow\ \zeta^2=-1-\zeta=-\fr{1}{2}\mp\fr{\sq3}{2}i \gr\ \Rightarrow\  \{\zeta^2=\overline{\zeta}\\|\zeta|=1\\|\zeta^2|=1                  
 
\bl|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=
 
\bl\ \ \ \ \ \ =|\alpha|^2+|\beta|^2+|\gamma\|^2+\alpha\overline\beta+\overline\alpha\beta+\alpha\overline\gamma+\overline\alpha\gamma+\beta\overline\gamma+\overline\beta\gamma
 
\bl|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=
 
\bl\ \ \ \ \ \ =|\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
 
\bl|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=
 
\bl\ \ \ \ \ \ \ =|\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
 
\re|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)=\[\ \\\ \\1+\zeta+\overline\zeta=0\\\ \\\ \]=
 
\re\ \ \ \ \ \ =3\(|\alpha|^2+|\beta|^2+|\gamma|^2\)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ :shifty: \ :shifty:

 


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