If `|z+1|=z+2(1+i),` find `zdot`

Explanation

Given that, `|z +1| = z + 2 ( + i) ... (i)`
`z = x + iy`
then, `|x + iy + 1| = x + iy + 2 (1 + i)`
`rArr |x + 1 + iy| = (x + 2) + i (y + 2)`
`rArr" sqrt((x + 1)^(2) + y^(2) = (x + 2) + i (y + 2))`
On squaring both sides, we get
`(x + 1)^(2) + y^(2) = (x + 2)^(2) +i ^(2)(y + 2)^(2) +2i (x + 2)(y +2)`
`rArr x^(2) + 2x + 1 + y^(2) = x^(2) + 4x + 4 - y^(2) - 4y - 4 +2i(x + 2)(y + 2)`
`rArr x^(2) + y^(2)+ 2x + 1 = x^(2) - y^(2) + 4x - 4y + 2i(x + 2)(y + 2)`
On comparing real and imaginary parts, we get
` x^(2) + y^(2) + 2x + 1 = x^(2) - y^(2) + 4x -4y`
`rArr 2y^(2) - 2x + 4y + 1 = 0 ...(ii)`
and `2(x + 2) (y + 2) =0`
`rArr x + 2 = 0 or y + 2 = 0 `
`x = - 2 or y = - 2 ...(iii)`
For `x = - 2, 2y^(2) + 4 + 4y + 1 = 0 [ using Eq. (ii)]`
`rArr 2y^(2) + 4y + 5 =`
`rArr 16 - 4 xx 2 xx 5 lt0 `
`:. "Discriminant" , D = b^(2) - 4ac lt 0 `
`rArr 2y^(2) + 4y + 5 "has no real roots"`.
For y = -2, `2(-)6(2) - 2x + 4 (-2) + 1 = 0 [ using Eq. (ii)]`
`rArr 8 - 2x - 8 + 1 = 0`
`rArr x = 1//2`
`:. " " z = x + iy = (1)/(2) - 2i`