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 nn {z :|z-1+i|=5}` is at most 2
D. If L has most than one elements, then L has infinitely many elements.

Answer

A, C, D

Explanation

Given `sz + bartz + z =0 , barz = x -iy" "(i)`
Taking conjugate, we get
`barz barz + bartz + barz = 0" "…(i)`
Adding (i) and (ii), we get
`(t + bars) barz + (bart + s)z+(r+barr) = 0`
Clearly, this is the equations of a straight line .
Now, eliminating `barz` form (i) and (ii) we,get
`sbarsz + rbars + rbars -tbartz - barrt =0`
`rArr (|s|^(2) - |t|^(2))z = barrt - rbars`
If `|S| ne|t|` then L has unique solution.
If `|s| = |t|` and `barrt = rbars`, then L has infinite solutions.
If `|s| = |t|` and `barrt ne rbars`, then L has no solutions.
Thus, L has either unique, infinite or no solution.
Also, line can intersect circle in maximum two points.
So, number of elements in L `nn{z:|z-1+i|=5}` is at most 2