fie z=a=bi, a, b∈R
atunci:
(a+bi)²+2a-2bi+2=0
a²-b²+2a+2+(2ab-2b)i=0+0i
a²-b²+2a+2=0
si
2b(a-1)=0
b=0 atunci a²-2a+2=0 a∉R
sau a=1
1-b²+2+2=0
b²=5 deci b=√5 sau b=-√5
deci z=1+i√5 sau z=1-i√5
se observa ca are 2 radacini complex conjugate
verificare (1+i√5)²+2-2i√5+2=1-5+2i√5+2-2i√5+2=0+0i
deci OK, numai verificam si pt conjugata
adevarat, bine rezolvat
