Let z=x+iy
2z+iiz−1=2(x+iy)i(x+iy)−1=2x+i(1+2y)−y−1+ix
=2x+i(1+2y)×−(1+y)=ix−(1+y)+ix×−(1+y)−ix
Given, Im=[2z+iiz−1]=−1
⇒−2x2−(1+2y)(1+y)[−(1+y)]2+x2=1
⇒−2x2−(1+y+2y+2y2)1+y2+2y+x2=−1
⇒−2x2−1−3y−2y2=−1−y2−2y−x2
⇒x2+y2+y=0
∴ locus of the P is therefore x2+y2+y=0.