If z is a complex number satisfying the equation `|z-(1+i)|^2=2` and `omega=2/z`, then the locus traced by `'omega'` in the complex plane is

Options

A. `(x-y+1)=0`
B. `x-y-1=0`
C. `x+y-1=0`
D. `x+y+1=0`

Answer

B

Explanation

Let `z=x+iy` and `omega=h+ik`. Then,
`|z-(1+i)|^(2)=2` and `omega=2/z`
`rArr |(x-1)+i(y-1)|^(2)` and `omega=2(x/(xp^(2)+y^(2))-(iy)/(x^(2)+y^(2)))`
`rArr (x-1)^(2)+(y-1)^(2)=2` and `h+ik = (2x)/(x^(2)+y^(2))+((-2y)i)/(x^(2)+y^(2))`
`rArr x^(2)+y^(2)=2(x+y)` and `h=(2x)/(x^(2)+y^(2))` and `(-2y)/(x^(2)+y^(2))`
`rArr 1=h-k`
Hence, the locus of (h,k) is `x-y-1=0`.