The equation Im(iz/(z−i))+1=0, z∈C, z≠i represents a part of a circle having radius equal to.

Im(iz/(z−i))+1=0
Writing z=x+iy, we have
Im(i(x+iy)/((x+iy)−i))+1=0
⇒Im(i(x+iy)/(x+i(y−1)))×(x−i(y−1))/(x−i(y−1)))+1=0
⇒Im(i(x+iy)(x−i(y−1))/(x2+(y−1)2))+1=0
⇒Im(i(x2+i(xy−x−y2+y)+xy−iy2+iy))/(x2+(y−1)2))+1=0
⇒Im(i(x2+xy−x−y2+y)+xy−iy2+iy)/(x2+(y−1)2))+1=0
⇒(x2−y2+xy−x+y)/(x2+(y−1)2)+1=0
⇒x2−y2+xy−x+y+x2+(y−1)2=0
⇒x2−y2+xy−x+y+x2+y2−2y+1=0
⇒2x2+xy−x−y+1=0
⇒2x2+xy−x−y+1=0
Let's consider this as a circle.
2x2+2y2+2gx+2fy+c=0
Comparing with the general equation of a circle x2+y2+2gx+2fy+c=0,
we have x2+y2+gx+fy+c=0
2x2+xy−x−y+1=0
2(x2+y2)+xy−x−y+1=0
This equation represents a part of a circle.
This equation is not of the form x2+y2+2gx+2fy+c=0.
Let the equation be
2x2+2y2−2y+1=0
2x2+2y2−2y+1=0
2x2+2(y2−y)+1=0
2x2+2(y2−y+1/4)−1/2+1=0
2x2+2(y−1/2)2+1/2=0
2x2+2(y−1/2)2=−1/2
x2+(y−1/2)2=−1/4
This equation does not represent a circle.
Im(iz/(z−i))+1=0
Let z=x+iy
Im(i(x+iy)/(x+i(y−1)))+1=0
Im(i(x+iy)(x−i(y−1))/(x2+(y−1)2))+1=0
(x2−y2+xy−x+y)/(x2+(y−1)2)+1=0
(x2−y2+xy−x+y)+x2+y2−2y+1=0
2x2+xy−x−y+1=0
This equation doesn't represent a circle.
If we complete the square, we get:
x^2 + y^2 - y/2 + 1/2 = 0
x^2 + (y - 1/4)^2 = -1/4 + 1/16 = -3/16
This isn't a circle. There must be a mistake in the question or solution provided.