If (z=x+i y, x, y R ) and the imaginary part of ( z -1 z -i ) is 1 , then the locus of (z ) is

Step-by-step solution

If (z=x+i y ), then ( aligned & z -1 z -i = x-i y-1 x-i y-i x+i(y+1) x+i(y+1) = & [x(x-1)+y(y+1)]+i[(y+1)(x-1)-x y] x^2+(y+1)^2 & Im ( z -1 z -i )= x y-y+x-1-x y x^2+(y+1)^2 =1 (given) & x^2+y^2-x+3 y+2=0,(x, y) (0,-1) aligned ) Hence, option (4) is correct.