Among the statements (S1) : The set \ z ∈ ℂ-\ -i\ :|z|=1. and z-i z+i is purely real contains exactly two elements, and (S2) : The set \ z ∈ ℂ-\ -1\ :|z|=1. and z-1 z+1 is purely imaginary contains infinitely many

Correct answer: C. only (S2) is correct aligned & S_1:|z|=1, z-i z+i = z +i z -i & (z-i)( z -i)=(z+i)( z +i) & |z|^2-i(z+ z )-1=|z|^2+i(z+ z )-1 & i(z+ z )=0 & z+ z =2 θ=0 θ=0 & z=0+0 i,|z| ≠ 1 & ~S_1: z-1 z+1 + z -1 z +1 =0 & (z-1)( z +1)+(z+1)( z -1)=0 & |z|^2+(z- z )-1+|z|^2+(z- z )-1=0 & |z|^2=1 aligned

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Complex Number

Among the statements (S1) : The set \ z ∈ ℂ-\ -i\ :|z|=1. and z-i...

Among the statements
(S1) : The set

\left\{\mathrm{z} \in \mathbb{C}-\{-\mathrm{i}\}:|\mathrm{z}|=1\right.\text{ and } \frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}\text{ is }\text{ purely }\text{ 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.

Options

Aboth are incorrect
Bonly (S1) is correct
Conly (S2) is correct
Dboth are correct

Correct answer

C. only (S2) is correct

Step-by-step solution

\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 \theta=0 \Rightarrow \cos \theta=0 \\ & \mathrm{z}=0+0 \mathrm{i},|\mathrm{z}| \neq 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}

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