Among the statements
(S1) : The set $\left\{\mathrm{z} \in \mathbb{C}-\{-\mathrm{i}\}:|\mathrm{z}|=1\right.$ and $\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}$ is purely real} contains exactly two elements, and (S2) : The set $\left\{\mathrm{z} \in \mathbb{C}-\{-1\}:|\mathrm{z}|=1\right.$ and $\frac{\mathrm{z}-1}{\mathrm{z}+1}$ is purely imaginary contains infinitely many elements.

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{S}_1:|\mathrm{z}|=1, \frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}=\frac{\overline{\mathrm{z}}+\mathrm{i}}{\overline{\mathrm{z}}-\mathrm{i}} \ & \Rightarrow(\mathrm{z}-\mathrm{i})(\overline{\mathrm{z}}-\mathrm{i})=(\mathrm{z}+\mathrm{i})(\overline{\mathrm{z}}+\mathrm{i}) \ & |\mathrm{z}|^2-\mathrm{i}(\mathrm{z}+\overline{\mathrm{z}})-1=|\mathrm{z}|^2+\mathrm{i}(\mathrm{z}+\overline{\mathrm{z}})-1 \ & \mathrm{i}(\mathrm{z}+\overline{\mathrm{z}})=0 \ & \mathrm{z}+\overline{\mathrm{z}}=2 \cos 	heta=0 \Rightarrow \cos 	heta=0 \ & \mathrm{z}=0+0 \mathrm{i},|\mathrm{z}| 
eq 1 \ & \mathrm{~S}_1: \frac{\mathrm{z}-1}{\mathrm{z}+1}+\frac{\overline{\mathrm{z}}-1}{\overline{\mathrm{z}}+1}=0 \ & (\mathrm{z}-1)(\overline{\mathrm{z}}+1)+(\mathrm{z}+1)(\overline{\mathrm{z}}-1)=0 \ & \Rightarrow|\mathrm{z}|^2+(\mathrm{z}-\overline{\mathrm{z}})-1+|\mathrm{z}|^2+(\mathrm{z}-\overline{\mathrm{z}})-1=0 \ & |\mathrm{z}|^2=1\end{aligned}$

Among the statements (S1) : The set $\left\{\mathrm{z} \in \mathbb{C}-\{-\mathrm{i}\}:|\mathrm{z}|=1\right.$…

Solution