Here z−iz+i=x+i(y−1x+i(y+1).x−i(y+1)x−i(y+1) =(x2+y2−1)+i(−2x)x2+(y+1)2 As z−iz+i is purely imaginary, we get x2+y2−1=0 ⇒ x2+y2=1 ⇒ z¯z=1.