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Question
JEE Main
Mathematics

Among the statements
(S1): The set $\{ z\in C-\{-i\} :|z| =1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\}$ contains exactly two elements, and
(S2): The set $\{z \in C-\{-1\}:|z| =1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\}$ contains infinitely many elements.

The correct answer is
only (S2) is correct

Analyzing Statement S1: Purely Real Condition

We examine statement S1: \( S_1 = \{ z\in \mathbb{C}-\{-i\} :|z| =1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\} \).

  1. The condition \( |z|=1 \) means \( z \) lies on the unit circle in the complex plane.
  2. Let \( w = \frac{z-i}{z+i} \). For \( w \) to be purely real, its imaginary part must be zero, i.e., \( \text{Im}(w) = 0 \).
  3. Representing \( z \) as \( x+iy \), after algebraic manipulation using \( |z|=1 \), we find \( w = \frac{-ix}{1+y} \).
  4. The imaginary part \( \frac{-x}{1+y} \) is zero only if \( x=0 \).
  5. Combining \( x=0 \) with \( |z|=1 \) (\( x^2+y^2=1 \)) yields \( y^2=1 \), meaning \( y=1 \) or \( y=-1 \). Potential solutions are \( z=i \) and \( z=-i \).
  6. The expression \( \frac{z-i}{z+i} \) is undefined for \( z=-i \). The set definition also explicitly excludes \( z=-i \).
  7. Thus, the only valid solution is \( z=i \). The set \( S_1 \) contains only one element, \( \{i\} \). Statement S1 is incorrect.

Analyzing Statement S2: Purely Imaginary Condition

We examine statement S2: \( S_2 = \{z \in \mathbb{C}-\{-1\}:|z| =1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\} \).

  1. The condition \( |z|=1 \) again means \( z \) lies on the unit circle.
  2. Let \( w = \frac{z-1}{z+1} \). For \( w \) to be purely imaginary, its real part must be zero, i.e., \( \text{Re}(w) = 0 \).
  3. Representing \( z \) as \( x+iy \), after algebraic manipulation using \( |z|=1 \), we find \( w = \frac{iy}{1+x} \).
  4. The real part of \( w \) is \( 0 \). This holds true provided the denominator \( 1+x \neq 0 \).
  5. The condition \( z \in \mathbb{C}-\{-1\} \) ensures \( x \neq -1 \).
  6. Therefore, any \( z \) on the unit circle \( |z|=1 \) except \( z=-1 \) satisfies the condition.
  7. This set is the unit circle excluding the point \( (-1,0) \), which contains infinitely many elements. Statement S2 is correct.

Conclusion

Statement S1 is incorrect because the set contains only one element. Statement S2 is correct because the set contains infinitely many elements. Therefore, only statement S2 is correct.

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