For all z  satisfying |z+1−i|=1  we have

|z+1−i|=1⇒|z−(−1)+i|=1       ∴  Locus of z is a circle whose centre is (−1,1)  and radius=1              The circle touches both the axes in the second quadrant      All points on this circle lie in the region +ve  Y-axis  corresponding to Arg z=π2  and-ve  X-axis  corresponds to Argz =π    ∴π2≤Arg Z≤π        Hence option ( 1)