The correct option is C 4
Let z=x+iy
As z2=¯¯¯z
⇒(x+iy)2=x−iy⇒x2−y2+2ixy=x−iy
Equality of complex numbers holds when
x2−y2=x, 2xy=−y⇒2xy=−y⇒y(2x+1)=0⇒y=0,x=−12
When y=0, then
x2−y2=x⇒x(x−1)=0⇒x=0,1
When x=−12, then
x2−y2=x⇒14−y2=−12⇒y2=34⇒y=±√32
Therefore,
(x,y)=(0,0),(1,0),(−12,√32),(−12,−√32)
Hence, there are 4 complex numbers satisfying the given equation.