If `z=x+iy` such that `|z+1|=|z-1|` and `arg((z-1)/(z+1))=pi/4` then

Options

A. `x^(2)-y^(2)-2x-1=0`
B. `x^(2)+y^(2)-2x-1=0`
C. `x^(2)+y^(2)-2y-1=0`
D. `x^(2)+y^(2)+2x-1=0`

Answer

C

Explanation

We have,
`(z-1)/(z+1) = ((x-1)+iy)/((x+1)+iy)=(x^(2)+y^(2)-1)/((x+1)^(2)+y^(2))`
`therefore "arg"(z-1)/(z+1)=pi/4`
`rArr tan^(-1)(2y)/(x^(2)+y^(2)-1)=pi/4`
`rArr (2y)/(x^(2)+y^(2)-1) = tanpi/4 rArr x^(2)+y^(2)-2y-1=0`