Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma0fcc: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma0fcc Solution: Let s, t, r be non-zero complex numbers and L be the set of…

Let s, t, r be non-zero complex numbers and L be the set of solutions z=x+iy(x,y∈R, i=−1)z = x + iy(x,y \in R,\,i = \sqrt { - 1} ) of the equation sz+tz‾+r=0sz + t\overline z + r = 0 where z‾\overline z = x −- iy. Then, which of the following statement(s) is(are) TRUE?

Multiple Choice 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∩{z:∣z−1+i∣=5}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

Step-by-Step Verified Solution

We have,

sz+tz‾+r=0sz + t\overline z + r = 0 ...(i)

On taking conjugate,

sz‾+t‾z+r‾=0\overline {sz} + \overline t z + \overline r = 0 ... (ii)

On solving Eqs. (i) and (ii), we get

z=r‾t−rs‾∣s∣2−∣t∣2z = {{\overline r t - r\overline s } \over {|s{|^2} - |t{|^2}}}

(a) For unique solutions of z

∣s∣2−∣t∣2≠0⇒∣s∣ ≠ ∣t∣|s{|^2} - |t{|^2} \ne 0 \Rightarrow |s|\, \ne \,|t|

It is true

(b) If |s| = |t|, then r‾t−rs‾{\overline r t - r\overline s } may or may not be zero. So, z may have no solutions.

∴\therefore L may be an empty set.

It is false.

(c) If elements of set L represents line, then this line and given circle intersect at maximum two point. Hence, it is true.

(d) In this case locus of z is a line, so L has infinite elements. Hence, it is true.

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