ben je vais vous montrer mon calcul detaillé, je crois que j'ai juste la ( j'espere ) :
( 1 + 2eiπ/32e^{iπ/3}2eiπ/3 + e2iπ/3e^{2iπ/3}e2iπ/3 ) / ( eiπ/3e^{iπ/3}eiπ/3 )
= eiπ/3e^{iπ/3}eiπ/3 ( 1 / (eiπ/3(e^{iπ/3}(eiπ/3) + 2 + eiπ/3e^{iπ/3}eiπ/3 ) / ( eiπ/3e^{iπ/3}eiπ/3 )
= 1 / ( eiπ/3e^{iπ/3}eiπ/3 ) + 2 + eiπ/3e^{iπ/3}eiπ/3
= e−iπ/3e^{-iπ/3}e−iπ/3 + 2 + eiπ/3e^{iπ/3}eiπ/3
= 2 + 1 = 3
Donc arg ( (ze-za)/(zb-za) ) ) = arg ( 3 ) = 0
les 3 points sont donc alignés !
j'ai juste ? ^^"