Find the range of real number `alpha` for which the equation `z+alpha|z-1|+2i=0` has a solution.

Explanation

Let `z = x + iy`
We have,
`z+ alpha|z-1| + 2i=0`
`rArr x + i(y+2) + alpha sqrt((x-1)^(2) + y^(2))=0`
Equating real and imaginary parts
`therefore y = - 2 and x + alpha sqrt((x-1)^(2)+4)=0`
`therefore x^(2) = alpha^(2) (x^(2)-2x +5)`
`therefore (1-alpha^(2))x^(2) + 2alpha^(2)x-5alpha^(2) = 0`
Since x is real,
`therefore D =b^(2) - 4ac ge 0`
`rArr 4alpha^(4) + 20alpha^(2)(1-alpha^(2)) ge 0`
`rArr 4alpha^(4) + 5alpha^(2) ge 0`
`rArr 4alpha^(2)(alpha^(2) - (5)/(4)) le 0`
`rArr (-sqrt(5))/(2) le alpha le(sqrt(5))/(2)`.