(D) Given $\left|\frac{z}{1+i}\right|=2$. Since $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $|z| = 2|1+i| = 2\sqrt{2}$. Substituting $z=x+iy$,we get $

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) Given $\left|\frac{z}{1+i}\right|=2$. Since $|1+i| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $|z| = 2|1+i| = 2\sqrt{2}$. Substituting $z=x+iy$,we get $|x+iy| = 2\sqrt{2}$. Squaring both sides,we get $x^2+y^2 = (2\sqrt{2})^2 = 8$. This represents a circle $x^2+y^2 = 8$ with centre $C \equiv (0,0)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Geometry of complex numbers

Difficult

If $\left|\frac{z}{1+i}\right|=2$,where $z=x+iy$ and $i=\sqrt{-1}$ represents a circle,then the centre $C$ and radius $r$ of the circle are:

MHT CET 2024 All MHT CET

A
$C \equiv(3,0), r=4$
B
$C \equiv(6,0), r=2$
C
$C \equiv(0,3), r=8$
D
$C \equiv(0,0), r=2\sqrt{2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

1 Complex Numbers and Quadratic Equations: Complex Numbers

Subtopic

Geometry of complex numbers

Previous Year

MHT CET 2024 Paper All MHT CET years →

If $P$ is a complex number whose modulus is $1$,then the equation $\left(\frac{1+iz}{1-iz}\right)^4=P$ has

EasyTS EAMCET 2019

View Solution

The area of the triangle whose vertices are complex numbers $z, iz, z + iz$ in the Argand diagram is

DifficultAIEEE 2012

View Solution

For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$,if $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$,then $|\alpha - \beta|$ is equal to $.............$.

DifficultJEE MAIN 2023

View Solution

If $z=x+iy$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,then the locus of $z$ is

MediumTS EAMCET 2002

View Solution

If $z = x + iy$ is a complex number,then the equation $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$ represents the

MediumMHT CET 2025

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $\left|\frac{z}{1+i}\right|=2$,where $z=x+iy$ and $i=\sqrt{-1}$ represents a circle,then the centre $C$ and radius $r$ of the circle are:

Similar Questions

If $P$ is a complex number whose modulus is $1$,then the equation $\left(\frac{1+iz}{1-iz}\right)^4=P$ has

The area of the triangle whose vertices are complex numbers $z, iz, z + iz$ in the Argand diagram is

For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$,if $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$,then $|\alpha - \beta|$ is equal to $.............$.

If $z=x+iy$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,then the locus of $z$ is

If $z = x + iy$ is a complex number,then the equation $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$ represents the

Vedclass Test Series

Exam Paper Generator

Online Exam Module