Let z and w be two complex numbers such that `|Z| <= 1, |w|<=1 and |z + iw| = |z-i bar w| = 2`

Options

A. 1 or I
B. I or -I
C. 1 or -1
D. I or -1

Answer

C

Explanation

Given `|z+I w|=|z-ibarw|=2`
`rArr |z-(-iw)|=|z-(bar iw)|=2`
`rArr |z-(-iw)|=|z-(-ibarW)|`
`therefore` z lies on the perpendicular bisector of the line joininig -iw and - `Ibarw ."since " - bar w` and y=0
Now `|z| le 1 rArr x^2 +0^2 le 1 rArr -1 le x le 1`
`therefore` z may take values given in opton ( C)
Alteranate Solution
|z+ iw | le |z| + |iw | = |z| + |w|
`le 1+1 =2`
`therefore |z+i w | le 2 `
`rArr` |z+iw|= 2 holds when
argz- arg i w =0
`rArr arg z-arg i w =0 `
` rArr arg z/(iw)=0`
` z/(iw) ` is purely real.
`rArr z/w ` is purely imaginary
Similarly when `|z-i bar w|=2 "then " z/w` is purely imaginary
Now , given relation`|z+iw|=|z-barw|=2`
Put w=i ,we get
`|z+ i^2|=|z+i^2|=2`
`rArr |z+1|=2 rArr z=1 [therefore |z| le 1]`
`therefore ` z=1 or -1 is the correct option.