Question

Let s, t, r be non-zero complex numbers and L be the set of solutions z= x+ iy (x, yeR,i=V-1) of the equation sz + tz + r=0, where z = x - iy. Then, which of the following statement(s) is (are) TRUE? (2018 Adv.) (a) If L has exactly one element, then sluit| (b) If sl=\t\, then L has infinitely many elements (c) The number of elements in Ln {z:/ 2-1+ i] =5} is at most 2 (d) If L has more than one element, then L has infinitely many elements

how to eliminate Z here
please answer fasr​

Answer 1

Answer:

If L has exactly one element, then ∣s∣=∣t∣

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

Given

sz+tz+r=0 ...(1)

On taking conjugate of the above equation  sz+tz+r=0 ...(2)

From (1) and (2) elliminating z

z(∣s∣2−∣t∣2)=rt−rs

(A) If ∣s∣=∣t∣ then z has unique value

(B) If ∣s∣=∣t∣ then rt−rs may or may not be zero. So L may be empty set.

(C) Locus of z is null set or singleton set or a line. In all cases it will intersect given circle at atmost two points.

(D) In this case locus of z is a line so L has infinite elements.