Question:

$(S_1)$: Let $|z| = 1$ and $\frac{z-i}{z+i}$ be purely real. Then $\frac{z-i}{z+i} = \overline{\left(\frac{z-i}{z+i}\right)}$ $\Rightarrow \frac{z-i}{z+i} = \frac{\bar{z}+i}{\bar{z}-i}$ $\Rightarrow (z-i)(\bar{z}-i) = (z+i)(\bar{z}+i)$ $\Rightarrow |z|^2 - i(z + \bar{z}) - 1 = |z|^2 + i(z + \bar{z}) - 1$ $\Rightarrow -i(z + \bar{z}) = i(z + \bar{z})$ $\Rightarrow z + \bar{z} = 0$ $\Rightarrow \text{Re}(z) = 0$ With $|z| = 1$ and $z = iy$, we get $|y| = 1$, so $z = i$ or $z = -i$. Since $z \neq -i$, we have only $z = i$. Thus $S_1$ contains exactly one element, not two. $(S_1)$ is incorrect. $(S_2)$: Let $|z| = 1$ and $\frac{z-1}{z+1}$ be purely imaginary. Then $\frac{z-1}{z+1} + \overline{\left(\frac{z-1}{z+1}\right)} = 0$ $\Rightarrow \frac{z-1}{z+1} + \frac{\bar{z}-1}{\bar{z}+1} = 0$ $\Rightarrow (z-1)(\bar{z}+1) + (\bar{z}-1)(z+1) = 0$ $\Rightarrow |z|^2 + (z-\bar{z}) - 1 + |z|^2 - (z-\bar{z}) - 1 = 0$ $\Rightarrow 2|z|^2 - 2 = 0$ $\Rightarrow |z|^2 = 1$ This is satisfied for all $z$ with $|z| = 1$. Thus $S_2$ contains infinitely many elements (the entire unit circle except $z = -1$). $(S_2)$ is correct.