Among the statements:(S1) : The set {z in mat | vedclass

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(C) For $(S1)$: Let $z = x+iy$. Since $|z|=1$,$x^2+y^2=1$. The condition that $\frac{z-i}{z+i}$ is purely real means $\frac{z-i}{z+i} = \overline{\lef

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only $(S2)$ is correct

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(C) For $(S1)$: Let $z = x+iy$. Since $|z|=1$,$x^2+y^2=1$. The condition that $\frac{z-i}{z+i}$ is purely real means $\frac{z-i}{z+i} = \overline{\left(\frac{z-i}{z+i}ight)} = \frac{\bar{z}+i}{\bar{z}-i}$.$(z-i)(\bar{z}-i) = (z+i)(\bar{z}+i) \Rightarrow |z|^2 - i(z+\bar{z}) - 1 = |z|^2 + i(z+\bar{z}) - 1$.This simplifies to $2i(z+\bar{z}) = 0$,so $z+\bar{z} = 2x = 0$,meaning $x=0$. Since $|z|=1$,$y^2=1$,so $z = \pm i$. However,$z 
eq -i$,so $z=i$ is the only solution. Thus,$(S1)$ is incorrect.For $(S2)$: Let $w = \frac{z-1}{z+1}$. The condition that $w$ is purely imaginary means $w + \bar{w} = 0$.$\frac{z-1}{z+1} + \frac{\bar{z}-1}{\bar{z}+1} = 0 \Rightarrow (z-1)(\bar{z}+1) + (z+1)(\bar{z}-1) = 0$.$|z|^2 + z - \bar{z} - 1 + |z|^2 - z + \bar{z} - 1 = 0 \Rightarrow 2|z|^2 - 2 = 0 \Rightarrow |z|^2 = 1$.Since $|z|=1$ is given,every $z$ on the unit circle (except $z=-1$) satisfies the condition. Thus,$(S2)$ is correct.

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