Question

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

• A. both are incorrect
• B. only (S1) is correct
• C. only (S2) is correct
• D. both are correct

Hint:

When dealing with conditions involving real and imaginary parts of complex functions, use algebraic manipulation to separate the real and imaginary components and analyze the conditions carefully.

Answer 1

The Correct Option is C

Step 1: Analysis of (S1).
The condition for \( \frac{z - i}{z + i} \) to be purely real is that its imaginary part is zero. Given \( z = x + iy \), algebraic manipulation reveals that this condition, along with \( |z| = 1 \), yields exactly two solutions. Consequently, the set \( \{ z \in \mathbb{C} - \{-i\} : |z| = 1 \text{ and } \frac{z - i}{z + i} \text{ is purely real} \} \) contains two elements, confirming statement (S1).

Step 2: Analysis of (S2).
The condition for \( \frac{z - 1}{z + 1} \) to be purely imaginary is that its real part is zero. Algebraic analysis demonstrates that for \( |z| = 1 \), there are infinitely many solutions to this equation, as infinitely many points on the unit circle satisfy the zero real part condition for the quotient. 
Therefore, statement (S2) is also correct. 

Step 3: Conclusion.
The correct conclusion is: \[ \text{Only (S2) is correct}. \]