Saved Bookmarks
| 1. |
Let z and w be two complex numbers such that `|Z| |
|
Answer» Correct Answer - C 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. |
|