(D) We know that $\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}$.Thus,$z = (e^{i\pi/6})^5 + (e^{-

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

(D) We know that $\frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{-i\pi/6}$.Thus,$z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5 = e^{i5\pi/6} + e^{-i5\pi/6}$.Using Euler's formula,$e^{i\theta} + e^{-i\theta} = 2\cos(\theta)$,we get $z = 2\cos(5\pi/6)$.Since $\cos(5\pi/6) = -\frac{\sqrt{3}}{2}$,we have $z = 2(-\frac{\sqrt{3}}{2}) = -\sqrt{3}$.Here,$R(z) = -\sqrt{3}$ and $I(z) = 0$.Therefore,$I(z) = 0$ is the correct statement.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

Difficult

Let $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

JEE MAIN 2019 All JEE MAIN

A
$R(z) = -3$
B
$R(z) > 0$ and $I(z) > 0$
C
$R(z) < 0$ and $I(z) > 0$
D
$I(z) = 0$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

4-1.Complex numbers

Subtopic

De Moivre's theorem and Roots of unity

Previous Year

JEE MAIN 2019 Paper All JEE MAIN years →

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

DifficultAP EAMCET 2010

View Solution

If $\omega$ represents a complex cube root of unity,then $\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=$

MediumAP EAMCET 2019

View Solution

One of the cube roots of unity is

Easy

View Solution

If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity,then the value of $(1-z_1)(1-z_2) \ldots (1-z_{n-1})$ is equal to

MediumAP EAMCET 2016

View Solution

If $\omega$ is an $n^{th}$ root of unity,other than unity,then the value of $1 + \omega + \omega^2 + ... + \omega^{n-1}$ is

Easy

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 $z = \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right)^5 + \left( \frac{\sqrt{3}}{2} - \frac{i}{2} \right)^5$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$,then:

Similar Questions

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

If $\omega$ represents a complex cube root of unity,then $\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=$

One of the cube roots of unity is

If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity,then the value of $(1-z_1)(1-z_2) \ldots (1-z_{n-1})$ is equal to

If $\omega$ is an $n^{th}$ root of unity,other than unity,then the value of $1 + \omega + \omega^2 + ... + \omega^{n-1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module