How do I solve JAMIA 2021 Mathematics questions?

Practice JAMIA 2021 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of JAMIA Mathematics questions?

JAMIA Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are JAMIA previous year papers available for free?

Yes. Mathem Solvex provides free access to JAMIA previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

+91 99359 65550 contact@maarula.in

Mathem SolvexGet the solution of every question

Home

Previous Year Questions

NIMCET CUET PG JAMIA MAH-CET AMU VITMEE

PYQ Paper Downloads Practice Paper

Latest Update

Resources

PYQ PDF Downloads Our Courses Test Series About Us

Home

Question Archive

Explore All PYQs NIMCET CUET PG JAMIA MAH-CET AMU VITMEE

Downloads & Practice

PDF Material Downloads Solve Practice Paper

Quick Links

Latest Updates About Us Our Premium Courses Contact Support

Preparing for Excellence...

MAARULA CLASSES

MCA Entrance Coaching

Quick Links

Home
Blog
About
Contact Us
Faculty

Exams

NIMCET
CUET PG
VITMEE
JAMIA
MAH-CET

Free Resources

Previous Year Papers
Free Mock Tests
Strategy Videos
Syllabus

Our Courses

Live Classes
Test Series
Comprehensive Notes

Download Maarula Mathem App

For Seamless learning experience

Google Play

Designed & Developed with ❤️ by Vivek Kumar | LinkedIn

Home PYQs

Updates Resources About

JAMIA 2021 Mathematics PYQ — The complex number which satisfies the condition lies on… | Mathem Solvex | Mathem Solvex

Home
Questions
JAMIA
Mathematics
Q. 2146

Tip:A–D to answerE for explanationV for videoS to reveal answer

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

The complex number $z$ which satisfies the condition $\frac{i + z}{i - z} = 1$ lies on

Choose the correct answer:

A.
Circle $x^{2} + y^{2} = 1$
B.
The $x$ -axis
(Correct Answer)
C.
The $y$ -axis
D.
The line $x + y = 1$

Correct Answer:

The $x$ -axis

Explanation

\left| \frac{i+z}{i-z} \right| = 1

\frac{|i+z|}{|i-z|} = 1

|i+z| = |i-z|

Let $z = x + iy$ :

|i + (x + iy)| = |i - (x + iy)|

|x + i(y+1)| = |-x + i(1-y)|

\sqrt{x^2 + (y+1)^2} = \sqrt{(-x)^2 + (1-y)^2}

x^2 + (y+1)^2 = x^2 + (1-y)^2

x^2 + y^2 + 2y + 1 = x^2 + 1 - 2y + y^2

2y = -2y

4y = 0 \implies y = 0

Since $y=0$ , the complex number $z$ lies on the Real Axis (x-axis).

Explanation

\left| \frac{i+z}{i-z} \right| = 1

\frac{|i+z|}{|i-z|} = 1

|i+z| = |i-z|

Let $z = x + iy$ :

|i + (x + iy)| = |i - (x + iy)|

|x + i(y+1)| = |-x + i(1-y)|

\sqrt{x^2 + (y+1)^2} = \sqrt{(-x)^2 + (1-y)^2}

x^2 + (y+1)^2 = x^2 + (1-y)^2

x^2 + y^2 + 2y + 1 = x^2 + 1 - 2y + y^2

2y = -2y

4y = 0 \implies y = 0

Since $y=0$ , the complex number $z$ lies on the Real Axis (x-axis).

Same Topic Questions

JAMIA

If …

Explore More

TopicAll "Complex Number" PYQs →SubjectMore Mathematics Questions →YearJAMIA 2021 Paper →All JAMIA Questions →📄 Download JAMIA PYQ Papers →📝 Preparation Tips & Articles →