Let `za n domega` be two complex numbers such that `|z|lt=1,|omega|lt=1a n d|z-iomega|=|z-i omega|=2,t h e nz` equals `1ori` b. `ior-i` c. `1or-1` d. `ior-1`

Options

A. `(2)/(3)`
B. `(sqrt(5))/(3)`
C. `(3)/(2)`
D. `(2sqrt(5))/(3)`

Answer

C

Explanation

We have `2=|z+iomega|le|z|+|omega|" "(therefore |z_(1)+z_(2)|le|z_(1)|+|z_(2)|)`
`therefore |z|+|omega|le2" "...(1)`
But it is given that `|z|le1 and |omega|le1`.
`rArr |z|+|omega|le2" "...(2)`
From (1) and (2),
`|z|=|omega|=1`
Also, `|z+iomega|=|z-(ibaromega)|`
`rArr |z-(-iomega)|=|z-(ibaromega)|`
This means that z lies on perpendicular bisector of the line segment joining `(-iomega) and (ibaromega)`, which is real axis, as `(-iomega) and (ibaromega)` are conjugate to each other.
For `z, Im(z) = 0`
If `z =x, " then " |z|le1`
`rArr x^(2)le1`
`rArr -1le x le 1`
Therefore, (3) is the correct option.