Let `s ,\ t ,\ r` be non-zero complex numbers and `L` be the set of solutions `z=x+i y\ \ (x ,\ y in RR,\ \ i=sqrt(-1))` of the equation `s z+t z +r=0` , where ` z =x-i y` . Then, which of the following statement(s) is (are) TRUE? If `L` has exactly one element, then `|s|!=|t|` (b) If `|s|=|t|` , then `L` has infinitely many elements (c) The number of elements in `Lnn{z :|z-1+i|=5}` is at most 2 (d) If `L` has more than one element, then `L` has infinitely many elements

Options

A. If L has exactly one element , then `|s| ne |t|`
B. If `|s|=|t|` then L has infinitely many elements
C. The number of elements in `L cap {z:|z-1+i|=5}` is at most 2
D. If L has more than one element , then L has infinitely many elements

Answer

A, C, D

Explanation

We have
`sz+t barz+r=0`
On taking conjugate
`bars bar z+ bari z+ bar r`=0
On solving Eqs (i) and (ii) we get
`z=(barrt-rbars)/(|s^2-|t|^2`
(a) For unique solitions of z
`|s|^2-|t|^2 ne 0 rArr |s| ne |t|`
It is true
(b) If |s|=|t| then `bar r t - r bar s ` may or may not be zero so,z may have no soluiton L may be an empty set . It is false
( c) If elements ofset L repersents line then this line and given circle intersect at maximum two point Hence it is true .