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Mathematics · Complex Numbers

JEE Main 2025 — 7 April, Morning Shift — Question 34

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II:

Statement I: The set {z∈C−{−i}:∣z∣=1\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right. and z−iz+i\frac{z-i}{z+i} is purely real }\} contains exactly two elements.

Statement II: The set {z∈C−{−1}:∣z∣=1\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right. and z−1z+1\frac{z-1}{z+1} is purely imaginary }\} contains infinitely many elements.

In the light of the above statements, choose the correct answer from the options given below :

  1. Option A:

    Both statement I and statement Il are correct.

  2. Option B:

    Statement I is correct and statement Il is incorrect.

  3. Option C:

    Statement I is incorrect and statement Il is correct.

    Correct
  4. Option D:

    Both statements 1 and statements ll are incorrect.

Answer: C

Step-by-step solution

z−iz+i=zˉ+izˉ−i\frac{z-i}{z+i}=\frac{\bar{z}+i}{\bar{z}-i}

=zzˉ−zˉ−iz−1=zzˉ+zi+izˉ−1=z \bar{z}-\bar{z}-i z-1=z \bar{z}+z i+i \bar{z}-1

=Z+Zˉ=0=Z+\bar{Z}=0

=2x=0=2 x=0

=x=0=x=0 ( yy-axis)

∣z∣=1|z|=1

∴z=i(z≠−i\therefore \quad z=i \quad(z \neq-i is given ))

Statement 1 is incorrect

z−iz+i+zˉ−1zˉ+1=0\frac{z-i}{z+i}+\frac{\bar{z}-1}{\bar{z}+1}=0

=zzˉ−zˉ+z−1+zzˉ−z+zˉ−1=0=z \bar{z}-\bar{z}+z-1+z \bar{z}-z+\bar{z}-1=0

=zzˉ=1=z \bar{z}=1 =∣z∣=1=|z|=1 Statement 2 is correct

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Morning Shift
Subject
Mathematics
Chapter
Complex Numbers
Topic
Geometry of Complex Numbers
Given below are two statements : one is labelled as Statement I and… | JEE Main 2025 PYQ with Solution · DhiX AI