Among the statements
(S1) : The set $\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right.$ and $\frac{z-i}{z+i}$ is purely real $\}$ contains exactly two elements, and
(S2) : The set $\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right.$ and $\frac{z-1}{z+1}$ is purely imaginary $\}$ contains infinitely many elements.
Solution
<p>$$\begin{aligned}
& \frac{z-i}{z+i}=\frac{\bar{z}+i}{\bar{z}-i} \\
& =z \bar{z}-i \bar{z}-i z-1=z \bar{z}+z i+i \bar{z}-1 \\
& =z+\bar{z}=0 \\
& =2 x=0 \\
& =x=0 \quad \text { (y-axis) }
\end{aligned}$$</p>
<p>$$\begin{aligned}
& |z|=1 \\
& \therefore \quad z=i \quad(z \neq-i \text { is given })
\end{aligned}$$</p>
<p>Statement 1 is incorrect</p>
<p>$$\begin{aligned}
& \frac{z-i}{z+i}+\frac{\bar{z}-1}{\bar{z}+1}=0 \\
& =z \bar{z}-\bar{z}+z-1+z \bar{z}-z+\bar{z}-1=0 \\
& =z \bar{z}=1 \\
& =|z|=1
\end{aligned}$$</p>
<p>Statement 2 is correct</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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