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?
(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?
Show Hint
- both are incorrect
- only (S1) is correct
- only (S2) is correct
- both are correct
The Correct Option is C
Approach Solution - 1
Step 1: Analyzing (S1).
Consider the equation \( \frac{z - i}{z + i} \) being purely real.
This means the imaginary part of \( \frac{z - i}{z + i} \) must be zero.
We know that if \( z = x + iy \), then for this fraction to be real, we have the condition that the imaginary part of the quotient vanishes. Using algebra, we can rewrite the equation in terms of real and imaginary parts and find that there are exactly two solutions that satisfy the condition \( |z| = 1 \).
Hence, \( \{ z \in \mathbb{C} - \{-i\} : |z| = 1 \text{ and } \frac{z - i}{z + i} \text{ is purely real} \} \) contains exactly two elements, so statement (S1) is correct.
Step 2: Analyzing (S2).
Consider the equation \( \frac{z - 1}{z + 1} \) being purely imaginary.
This means the real part of \( \frac{z - 1}{z + 1} \) must be zero. Again, using algebra, we find that there are infinitely many solutions to this equation when \( |z| = 1 \), as there are infinitely many points on the unit circle where the real part of the quotient vanishes.
Therefore, statement (S2) is also correct.
Step 3: Conclusion.
Thus, the correct answer is: \[ \text{Only (S2) is correct}. \]
Approach Solution -2
Given: \[ S_1 : |z| = 1, \quad \frac{z - i}{z + i} = \frac{\bar{z} + i}{\bar{z} - i} \] \[ \Rightarrow (z - i)(\bar{z} - i) = (z + i)(\bar{z} + i) \] \[ |z|^2 - i(z + \bar{z}) - 1 = |z|^2 + i(z + \bar{z}) - 1 \] \[ i(z + \bar{z}) = 0 \] \[ z + \bar{z} = 2\cos\theta = 0 \Rightarrow \cos\theta = 0 \] \[ z = 0 + 0i, \quad |z| \ne 1 \] Now consider: \[ S_1 : \frac{z - 1}{z + 1} + \frac{\bar{z} - 1}{\bar{z} + 1} = 0 \] \[ (z - 1)(\bar{z} + 1) + (z + 1)(\bar{z} - 1) = 0 \] \[ \Rightarrow |z|^2 + (z - \bar{z}) - 1 + |z|^2 + (\bar{z} - z) - 1 = 0 \] \[ \Rightarrow 2|z|^2 - 2 = 0 \] \[ \boxed{|z|^2 = 1} \]
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