(C) Given $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$.Multiplying by $z_1 z_2$,we get $z_1^2 + z_2^2 = z_1 z_2$.This can be rewritten as $z_1^2 + z_2^2 -

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

$z_1, z_2$ and the origin form an equilateral triangle

(C) Given $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$.Multiplying by $z_1 z_2$,we get $z_1^2 + z_2^2 = z_1 z_2$.This can be rewritten as $z_1^2 + z_2^2 - z_1 z_2 = 0$.Let $z_3 = 0$ be the origin.The condition for three points $z_1, z_2, z_3$ to form an equilateral triangle is $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.Substituting $z_3 = 0$,we get $z_1^2 + z_2^2 = z_1 z_2$,which matches our given equation.Therefore,$z_1, z_2$ and the origin form an equilateral triangle.

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

Medium

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then

A
$z_1, z_2$ are collinear
B
$z_1, z_2$ and the origin form a right-angled triangle
C
$z_1, z_2$ and the origin form an equilateral triangle
D
None of these

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

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg(z_{1}-z_{2})=\frac{\pi}{4}$ and $z_{1}, z_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to ..... .

DifficultJEE MAIN 2021

View Solution

Let $\arg(z)$ represent the principal argument of the complex number $z$. The curves $|z|=3$ and $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ intersect:

MediumJEE MAIN 2022

View Solution

Let $S = \{z \in \mathbb{C} : |z-1|=1 \text{ and } (\sqrt{2}-1)(z+\bar{z}) - i(z-\bar{z}) = 2\sqrt{2}\}$. Let $z_1, z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $|\sqrt{2}z_1 - z_2|^2$ equals:

DifficultJEE MAIN 2024

View Solution

If the eight vertices of a regular octagon are given by the complex numbers $z_j = \frac{1}{x_j - 2i}$ for $j = 1, 2, \dots, 8$,where $x_j$ are the roots of $x^8 - 1 = 0$,then the radius of the circumcircle of the octagon is

DifficultTS EAMCET 2025

View Solution

The equation $z \bar{z} + (2 - 3i) z + (2 + 3i) \bar{z} + 4 = 0$ represents a circle of radius (in $\text{ units}$)

EasyWBJEE 2020

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

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then

Similar Questions

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg(z_{1}-z_{2})=\frac{\pi}{4}$ and $z_{1}, z_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to ..... .

Let $\arg(z)$ represent the principal argument of the complex number $z$. The curves $|z|=3$ and $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ intersect:

Let $S = \{z \in \mathbb{C} : |z-1|=1 \text{ and } (\sqrt{2}-1)(z+\bar{z}) - i(z-\bar{z}) = 2\sqrt{2}\}$. Let $z_1, z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $|\sqrt{2}z_1 - z_2|^2$ equals:

If the eight vertices of a regular octagon are given by the complex numbers $z_j = \frac{1}{x_j - 2i}$ for $j = 1, 2, \dots, 8$,where $x_j$ are the roots of $x^8 - 1 = 0$,then the radius of the circumcircle of the octagon is

The equation $z \bar{z} + (2 - 3i) z + (2 + 3i) \bar{z} + 4 = 0$ represents a circle of radius (in $\text{ units}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module