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.
We examine statement S1: \( S_1 = \{ z\in \mathbb{C}-\{-i\} :|z| =1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\} \).
We examine statement S2: \( S_2 = \{z \in \mathbb{C}-\{-1\}:|z| =1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\} \).
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.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.