All complex numbers 'z' which satisfy the relation `|z-|z+1||=|z+|z-1||` on the complex plane lie on the

Options

A. `y=x`
B. `y=-x`
C. circle `x^(2)+y^(2)=1`
D. line `x=0` or on a line segment joining `(-1,0) to `(1,0)`

Answer

D

Explanation

`(d)` Given `|z-|z+1||^(2)=|z+|z-1||^(2)`
`:. (z-|z+1|)(barz-|z+1|)=(z+|z-1|)(barz+|z-1|)`
`zbarz-z|z+1|-barz|z+1|+|z+1|^(2)`
`=zbarz+z|z-1|+barz|z-1|+|z-1|^(2)`
`|z+1|^(2)-|z-1|^(2)=(z+barz)[|z-1|+|z+1|]`
`(z+1)(barz+1)-(z-1)(barz-1)=(z+barz)[|z-1|+|z+1|]`
`(zbarz+z+barz+1)-(zbarz-z-barz+1)=(z+barz)[|z-1|+|z+1|]`
`2(z+barz)=(z+barz)[|z+1|+|z-1|]`
`(z+barz)[|z+1|+|z-1|-2]=0`
`implies` Either `z+barz=0implies z` is purely imaginary
`impliesz` lies on `y` axis `impliesx=0`
or `|z+1|+|z-1|=2`
`impliesz` lies on the line segment joining `(-1,0)` and `(1,0)`