Among the statements (S1) : The set contains exactly two elements, and (S2) : The set contains infinitely many elements. [2025]

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Solution

Q.
Among the statements

(S1) : The set ${z \in C - {i} : | z | = 1 and \frac{z - i}{z + i} is purely real}$ contains exactly two elements, and

(S2) : The set ${z \in C - {- 1} : | z | = 1 and \frac{z - 1}{z + 1} is purely imagin ary}$ contains infinitely many elements. [2025]

1 only (S2) is correct

2 both are incorrect

3 only (S1) is correct

4 both are correct

Ans.
(1)

(S1) : $\frac{z - i}{z + i} = \frac{\bar{z} + i}{\bar{z} - i}$ [ $∵$ Purely real]

$\Rightarrow z \bar{z} - i \bar{z} - i z - 1 = z \bar{z} + z i + i \bar{z} - 1$

$\Rightarrow z + \bar{z} = 0$

$\Rightarrow 2 x = 0$ [ $∵$ Purely real]

$\Rightarrow x = 0$

Also, |z| = 1

$∴ z = i$ ( $∵ z \neq - i i s g i v e n$ )

(S1) is incorrect.

(S2) : $\frac{z - 1}{z + 1} + \frac{\bar{z} - 1}{\bar{z} + 1} = 0$ [ $∵$ Purely imaginary]

$\Rightarrow z \bar{z} - \bar{z} + z - 1 + z \bar{z} - z + \bar{z} - 1 = 0 \Rightarrow z \bar{z} = 1 \Rightarrow$ $| z | = 1$ ,

which represents a circle with radius 1 and centre (0, 0).

(S2) is correct.

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