NCERT Solutions for Exercise 5.2 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

# NCERT Solutions for Exercise 5.2 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

Edited By Vishal kumar | Updated on Nov 06, 2023 07:10 AM IST

## NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations Exercise 5.2- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 5: Complex Numbers and Quadratic Equations Exercise 5.2- NCERT solutions for exercise 5.2 Class 11 Maths Chapter 5 discuss the argand plane and polar representation of complex numbers. Class 11 maths ex 5.2 discuss arguments and modulus of complex numbers. Also, the NCERT Solutions for Class 11 Maths chapter 5 exercise 5.2 gives numerically related to conversion to polar form. To solve Class 11 Maths chapter 5 exercise 5.2 students require the knowledge of representation of complex numbers graphically. The modulus and argument of a complex number are understood graphically and is used to solve Class 11th Maths chapter 5 exercise 5.2.

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Other than the NCERT book Class 11 Maths chapter 5 exercise 5.2, the following exercises are also present in the NCERT syllabus Class 11 Maths, which are prepared by subject matter experts at Careers360, are highly detailed and presented in an easy-to-understand language. Additionally, these 11th class maths exercise 5.2 answers are available in PDF format, allowing students to download them for offline use at their convenience, without incurring any costs. This class 11 maths ex 5.2 solution resource is designed to help students improve their understanding of mathematics effectively.

### Access Complex Numbers And Quadratic Equations Class 11 Chapter 5-Exercise: 5.2

Given the problem is
$z=-1-i\sqrt{3}$
Now, let
$r\cos \theta = - 1 \ \ \ and \ \ \ r\sin \theta = -\sqrt3$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-\sqrt3)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+3$
$r^2 =4$
$r= 2$ $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta = -1 \ \ \ and \ \ \ 2\sin \theta = -\sqrt3$
$\cos \theta = -\frac{1}{2} \ \ \ and \ \ \ \sin \theta = -\frac{\sqrt3}{2}$
Since, both the values of $\cos \theta \ and \ \sin \theta$ is negative and we know that they are negative in III quadrant
Therefore,
Argument = $-\left ( \pi - \frac{\pi}{3} \right )= - \frac{2\pi}{3}$
Therefore, the argument is

$- \frac{2\pi}{3}$

Given the problem is
$z=-\sqrt{3}+i$
Now, let
$r\cos \theta = - \sqrt3 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-\sqrt3)^2+(1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+3$
$r^2 =4$
$r= 2$ $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta = -\sqrt3 \ \ \ and \ \ \ 2\sin \theta = 1$
$\cos \theta = -\frac{\sqrt3}{2} \ \ \ and \ \ \ \sin \theta =\frac{1}{2}$
Since values of $\cos \theta$ is negative and value $\sin \theta$ is positive and we know that this is the case in II quadrant
Therefore,
Argument = $\left ( \pi - \frac{\pi}{6} \right )= \frac{5\pi}{6}$
Therefore, the argument is

$\frac{5\pi}{6}$

Given problem is
$z=1-i$
Now, let
$r\cos \theta = 1 \ \ \ and \ \ \ r\sin \theta = -1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (1)^2+(-1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$ $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = 1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1$
$\cos \theta = \frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}$
Since values of $\sin \theta$ is negative and value $\cos \theta$ is positive and we know that this is the case in the IV quadrant
Therefore,
$\theta = -\frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ IV \ quadrant)$
Therefore,
$1-i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( -\frac{\pi}{4} \right ) +i\sqrt2\sin \left ( -\frac{\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )$

Therefore, the required polar form is $\sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )$

Given the problem is
$z=-1+i$
Now, let
$r\cos \theta = -1 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (1)^2+(-1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$ $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = -1 \ \ \ and \ \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =\frac{1}{\sqrt2}$
Since values of $\cos \theta$ is negative and value $\sin \theta$ is positive and we know that this is the case in II quadrant
Therefore,
$\theta =\left ( \pi - \frac{\pi}{4} \right )= \frac{3\pi}{4}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
Therefore,
$-1+i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( \frac{3\pi}{4} \right ) +i\sqrt2\sin \left ( \frac{3\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( \frac{3\pi}{4} \right ) +i\sin \left ( \frac{3\pi}{4} \right ) \right )$

Therefore, the required polar form is $\sqrt2\left ( \cos \left ( \frac{3\pi}{4} \right ) +i\sin \left ( \frac{3\pi}{4} \right ) \right )$

Given problem is
$z=-1-i$
Now, let
$r\cos \theta = -1 \ \ \ and \ \ \ r\sin \theta = -1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$ &nbsnbsp; $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = -1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}$
Since values of both $\cos \theta$ and $\sin \theta$ is negative and we know that this is the case in III quadrant
Therefore,
$\theta =-\left ( \pi - \frac{\pi}{4} \right )= -\frac{3\pi}{4}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ III \ quadrant)$
Therefore,
$-1-i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( -\frac{3\pi}{4} \right ) +i\sqrt2\sin \left (- \frac{3\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( -\frac{3\pi}{4} \right ) +i\sin \left ( -\frac{3\pi}{4} \right ) \right )$

Therefore, the required polar form is $\sqrt2\left ( \cos \left (- \frac{3\pi}{4} \right ) +i\sin \left (- \frac{3\pi}{4} \right ) \right )$

Given problem is
$z=-3$
Now, let
$r\cos \theta = -3 \ \ \ and \ \ \ r\sin \theta = 0$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-3)^2+(0)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 9+0$
$r^2 =9$
$r= 3$ $(\because r > 0)$
Therefore, the modulus is 3
Now,
$3\cos \theta =- 3 \ \ \ and \ \ \ 3\sin \theta = 0$
$\cos \theta = -1\ \ \ and \ \ \ \sin \theta =0$
Since values of $\cos \theta$ is negative and $\sin \theta$ is Positive and we know that this is the case in II quadrant
Therefore,
$\theta =\pi\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
Therefore,
$-3= r\cos \theta +ir\sin \theta$
$= 3\cos \left (\pi \right ) +i3\sin \left (\pi \right )$
$= 3\left ( \cos \pi +i\sin\pi \right )$

Therefore, the required polar form is $3\left ( \cos \pi +i\sin\pi \right )$

Given problem is
$z=\sqrt3+i$
Now, let
$r\cos \theta = \sqrt3 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (\sqrt3)^2+(1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 3+1$
$r^2 =4$
$r= 2$ $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta =\sqrt3 \ \ \ and \ \ \ 2\sin \theta = 1$
$\cos \theta = \frac{\sqrt3}{2}\ \ \ and \ \ \ \sin \theta =\frac{1}{2}$
Since values of Both $\cos \theta$ and $\sin \theta$ is Positive and we know that this is the case in I quadrant
Therefore,
$\theta =\frac{\pi}{6}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)$
Therefore,
$\sqrt3+i= r\cos \theta +ir\sin \theta$
$= 2\cos \left (\frac{\pi}{6} \right ) +i2\sin \left (\frac{\pi}{6} \right )$
$= 2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

Therefore, the required polar form is $2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

Given problem is
$z = i$
Now, let
$r\cos \theta = 0 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (0)^2+(1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 0+1$
$r^2 =1$
$r= 1$ $(\because r > 0)$
Therefore, the modulus is 1
Now,
$1\cos \theta =0 \ \ \ and \ \ \ 1\sin \theta = 1$
$\cos \theta =0\ \ \ and \ \ \ \sin \theta =1$
Since values of Both $\cos \theta$ and $\sin \theta$ is Positive and we know that this is the case in I quadrant
Therefore,
$\theta =\frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)$
Therefore,
$i= r\cos \theta +ir\sin \theta$
$= 1\cos \left (\frac{\pi}{2} \right ) +i1\sin \left (\frac{\pi}{2} \right )$
$= \cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

Therefore, the required polar form is $\cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

## More About NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2

A complex number is represented in the argand plane using two-axis that is the real and imaginary axis. The real part of a complex number is represented by the real axis and the imaginary part by the imaginary axis. The complex number a+ib can be written as ordered pair (a,b) where a is the real part and b is the imaginary part. These concepts are discussed prior to Exercise 5.2 Class 11 Maths. Questions related to the concepts of geometric form and polar form of complex numbers discussed in Class 11 Maths chapter 5 exercise 5.2.

Benefits of NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2

• Class 11 Maths chapter 5 exercise 5.2 will give practice questions related to modulus and arguments and polar representations of complex numbers

• Practising the exercise 5.2 Class 11 Maths will give a good understanding of the concepts discussed in the NCERT Class 11 Maths book topics after the first exercise and before the Class 11th Maths chapter 5 exercise 5.2

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## Key Features for NCERT Solutions of Exercise 5.2 class 11 maths

1. Comprehensive Solutions: Detailed, step-by-step 11th class maths exercise 5.2 answers provided for each exercise to facilitate a strong grasp of mathematical concepts.

2. Clarity and Precision: Ex 5.2 class 11 solutions are presented in a clear and accurate manner, ensuring students can confidently prepare for their exams and deepen their understanding.

3. Curriculum Aligned: These class 11 maths ex 5.2 solutions are closely aligned with the NCERT curriculum, covering topics and concepts as per the official syllabus.

4. Practice Opportunities: Class 11 ex 5.2 offers a diverse set of practice problems to help students enhance their problem-solving skills and reinforce their knowledge.

5. Free Accessibility: Typically available at no cost, ensuring wide accessibility to students.

## Subject Wise NCERT Exampler Solutions

1. How many questions are solved in the NCERT solutions for Class 11 Maths chapter 5 exercise 5.2?

Eight questions are solved in exercise 5.2 Class 11 Maths

2. What concepts are required to cover while solving Class 11 Maths chapter 5 exercise 5.2?

The concepts of argand plane, modulus and arguments and polar form of complex numbers should be covered before solving Class 11 Maths chapter 5 exercise 5.2

3. Write the ordered pair of the complex number i.

i =0+i can be written as (0,1) as ordered pair

4. What is the real and imaginary part of -2i+4?

Real part=4 and imaginary part=-2

5. What is the argument of i?

The argument of i is pi/2.

6. What is the multiplicative inverse of i?

The multiplicative inverse of i is 1/i=i^4/i=i^3=-i

7. What is the multiplicative inverse of -i?

The multiplicative inverse of -i is 1/-i=i^4/-i=-i^3=i

8. Find the value of i^{20}?

i^{20}=(i^4)^5=15=1

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