Let s ,   t ,   r be non-zero complex numbers and L be the set of solutions z = x + i y   x ,…

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+tz-+r=0, where z-=x-iy. Then, which of the following statement(s) is (are) TRUE?
  1. If L has exactly one element, then s≠t
  2. If s=t, then L has infinitely many elements
  3. The number of elements in L∩z:z-1+i=5 is at most 2
  4. If L has more than one element, then L has infinitely many elements

Solution

Given
sz+tz-+r=0 ...(i)
On taking conjugate s-z-+t-z+r-=0 ...(ii)
zs2-t2=r-t-rs-
1 If s≠t then z has unique value
2 If s=t then r-t-rs- may or may not be zero so L may be empty set
3 Locus of z is null set or singleton set or a line in all cases it will intersect given circle at most two points.
4 In this case locus of z is a line so L has infinite elements

Asked in: JEE Advanced 2018 (Paper 2)

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