Question

Let s, t, r be non-zero complex numbers and L be the set of solutions z = x + iy (x, y∈ R, i = √-1 ) of the equation sz +bar tz + r = 0 , where bar z =x - iy . Then, which of the following statement(s) is (are) TRUE ?
(A) 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 L ∩ {z : |z -1 + i| = 5} is at most 2
(D) If L has more than one element, then L has infinitely many elements

Answer 1

Answer:A,C,D

Solution:

(B) s = t = i and r = i => L has no element
(C) Adding both equations (s+bar t)z+(bar s+t)bar z+(r+bar r)= 0 which is a line if s +bar t ≠0
If s+bar t =0 => t = -bar s , sz - bar sz +r =0
Which either represents no points or a straight line
In either case, number of points of intersection of given circle and L cannot be more than 2
(D) If α and β both satisfy the given equation then any number of form λα + (1 – λ)β satisfies the given equation where λ ∈ R