z=a+bi =>z²=a²+2abi-b²
z(barat)=a-bi=>2z(bar)=2a-2bi
Ecuatia devine
(a²-b²+2a)-2bi*(a-1)=-1 =>
(a²-b²+2a)-2bi*(a-1)=-1 => sau a-1=0 , a=1 sau b=0
caz 1 a=1 atunci prima paranteza, partea reala devine
1-b²+2= - 1 =>b²=4 =>b= +/-2
cazul b=0 conduce la a²+2a+1=0 (a+1)²=0 => a=1 deci z=1 fals 1∉C
Deci z1=1-2i si
z2=1+2i
