\z\=\z-1\=>1)z=z-1=>0=-1-->F
2)-z=z-1=>2z-1=0=>2z=1=>z=0,5
3)z=-z+1=>2z=1=>z=0,5
4)-z=-z+1=>0=-1-->F
obs:nu mai e nevoie sa luam cazurile 3 si 4, ele dand acelasi rezultat ca si 1 si 2
\z\=\z^-1\=>1)z=z^-1=>z=1/z=>0,5=1/0,5=>0,5=2-->F
2)-z=z^-1=>-z=1/z=>-0,5=1/0,5=>-0,5=2-->F
=>z apartine multimii vide
