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NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations Exercise 4.1

Last updated date: 16th Jul 2024
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NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 - FREE PDF Download

Class 11 Maths Chapter 4 Exercise 4.1 Solutions PDF  focuses on complex numbers, a fundamental concept in mathematics that extends the idea of real numbers. This exercise introduces you to the basic properties and operations involving complex numbers, including addition, subtraction, multiplication, and division. This exercise will also help you understand the graphical representation of complex numbers on the complex plane Students can download the revised Class 11 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 4 Exercise 4.1 Class 11 | Vedantu
3. Formulas Used in Class 11 Chapter 4 Exercise 4.1
4. Access NCERT Solutions for Maths Class 11 Chapter 4 - Complex Numbers and Quadratic Equations
4.1Exercise 4.1
5. Class 11 Maths Chapter 4: Exercises Breakdown
6. CBSE Class 11 Maths Chapter 4 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Class 11 Chapter 4 Maths Exercise 4.1 Solutions are aligned with the updated CBSE guidelines for Class 11, ensuring students are well-prepared for exams. Access the Class 11 Maths Syllabus here.

Glance on NCERT Solutions Maths Chapter 4 Exercise 4.1 Class 11 | Vedantu

• NCERT Solution for Class 11 Maths Chapter 4 Exercise 4.1 Solutions Complex Numbers covers the topics Complex Numbers, Algebra of Complex Numbers, The Modulus, and the Conjugate of a Complex Number.

• Complex Numbers go beyond real numbers (like 1, 2, or -3) and incorporate the imaginary unit "i," defined as the square root of -1. So, a complex number looks like z = a + bi, where a and b are real numbers and bi represents the imaginary part.

• Complex numbers can be added, subtracted, multiplied, and divided. NCERT solutions will guide you through these operations, explaining how to handle the imaginary unit "i" during calculations.

• The modulus (|z|) of a complex number z = a + bi represents its distance from zero on the complex plane.

• Two complex numbers are equal if and only if their real and imaginary parts are equal. (z1 = z2 if and only if a1 = a2 and b1 = b2)

• The conjugate (z̅) of a complex number is simply its mirror image across the real axis, where the imaginary part flips the sign (z̅ = a - bi). Understanding these properties is crucial for working with complex numbers.

• Class 11 Maths Chapter 4 Exercise 4.1 Solutions PDF covers 14 fully solved questions and solutions

Formulas Used in Class 11 Chapter 4 Exercise 4.1

• Modulus Formula: The modulus (distance from zero) of a complex number z = a + bi is calculated as: |z| = √(a² + b²)

Competitive Exams after 12th Science

Access NCERT Solutions for Maths Class 11 Chapter 4 - Complex Numbers and Quadratic Equations

Exercise 4.1

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib

1. $(5i)\left( - \dfrac{3}{5}i \right)$

Ans:

$\Rightarrow$ $(5i)\left( \dfrac{-3}{5}i \right)=-5\times \dfrac{3}{5}\times i\times i$

$=-3{{i}^{2}}$

$=-3(-1)$       $\left[ {{i}^{2}}=-1 \right]$

$=3$

2. ${{i}^{9}}+{{i}^{19}}$

Ans:

$\Rightarrow $${{i}^{9}}+{{i}^{19}}={{i}^{4\times 2+1}}+{{i}^{4\times 4+3}} ={{\left( {{i}^{4}} \right)}^{2}}\cdot i+{{\left( {{i}^{4}} \right)}^{4}}\cdot {{i}^{3}} =1\times i+1\times (-i)\quad \left[ {{i}^{4}}=1,{{i}^{3}}=-i \right] =i+(-i) =0 3. {{i}^{-39}} Ans: \Rightarrow$${{i}^{ - 39}}$$= {{i}^{- 4 \times 9 - 3}} = {{\left( {{i}^{4}} \right)}^{- 9}}.{{i}^{- 3}} = {{\left( 1 \right)}^{-9}}.{{i}^{- 3}} \left[ {{i}^{4}} = 1 \right] = \dfrac{1}{{{i}^{3}}} = \dfrac{1}{- i} \left[ {{i}^{3}} = - i \right] = \dfrac{- 1}{i} \times \dfrac{i}{i}$$= \dfrac{- i}{{{i}^{2}}} = \dfrac{- i}{- 1} = i \left[ {{i}^{2}} = - 1 \right]$

4. $3\left( 7 + i7 \right) + i\left( 7 + i7 \right)$

Ans:

$\Rightarrow $$3\left( 7 + i7 \right) + i\left( 7 + i7 \right) = 21 + 21i +7i +7{{i}^{2}} = 21 + 28i +7\times \left( - 1 \right) \left[ \because {{i}^{2}} = - 1 \right] = 14 + 28i 5. $\left( 1 - i \right) - \left( -1 + 6i \right)$ Ans: $\Rightarrow \left( 1 - i \right) - \left( - 1 + i6 \right)$ $= 1 - i + 1 - 6i$ $=2 - 7i$ 6. \left( \dfrac{1}{5} + i\dfrac{2}{5} \right) - \left( 4 + i\dfrac{5}{2} \right). Ans: \Rightarrow \left( \dfrac{1}{5} + i\dfrac{2}{5} \right) - \left( 4 + i\dfrac{5}{2} \right) = \dfrac{1}{5} + \dfrac{2}{5}i - 4 - \dfrac{5}{2}i = \left( \dfrac{1}{5} - 4 \right) + \left( \dfrac{2}{5} - \dfrac{5}{2} \right)i = \left( \dfrac{1 - \left( 4 \times 5 \right)}{5} \right) + \left( \dfrac{\left( 2 \times 2 \right) - \left( 5 \times 5 \right)}{5 \times 2} \right)i = - \left( \dfrac{1 - 20}{5} \right) + \left( \dfrac{4 - 25}{10} \right)i = - \dfrac{19}{5} + \left( \dfrac{- 21}{10} \right)i = - \dfrac{19}{5} - \dfrac{21}{10}i 7. \left[ \left( \dfrac{1}{3} + i\dfrac{7}{3} \right) + \left( 4 + i\dfrac{1}{3} \right) - \left( - \dfrac{4}{3} + i \right) \right]. Ans: \Rightarrow$$\left[ \left( \dfrac{1}{3} + i\dfrac{7}{3} \right) + \left( 4 + i\dfrac{1}{3} \right) - \left( - \dfrac{4}{3} + i \right) \right]$

$\dfrac{1}{3} + \dfrac{7}{3}i + 4 + \dfrac{1}{3}i + \dfrac{4}{3} - i$

$=\left( \dfrac{1}{3} + 4 + \dfrac{4}{3} \right) + \left( \dfrac{7}{3} + \dfrac{1}{3} - 1 \right)i$

$= \left( \dfrac{1 + \left( 4 \times 3 \right) + 4}{3} \right) + \left( \dfrac{7 + 1 - \left( 1 \times 3 \right)}{3} \right)i$

$= \left( \dfrac{1 + 12 + 4}{3} \right) + \left( \dfrac{7 + 1 - 3}{3} \right)i$

$a + ib:$$= \dfrac{17}{3} + i\dfrac{5}{3}$

8. ${{\left( 1 - i \right)}^{4}}$

Ans:

$\Rightarrow$${{\left( 1 - i \right)}^{4}} = {{\left[ {{\left( 1 - i \right)}^{2}} \right]}^{2}}$

$= {{\left[ {{1}^{2}} + {{i}^{2}} -2i \right]}^{2}}$

${{\left( 1 - i \right)}^{4}} = {{\left[ {{\left( 1 - i \right)}^{2}} \right]}^{2}}$

$={{\left[ 1 - 1 - 2i \right]}^{2}}$

$={{\left( 2i \right)}^{2}}$

$=4{{i}^{2}}$

$=-4 \left[ {{i}^{2}} = - 1 \right]$

$a + ib = -4 + 0i$

9. ${{\left( \dfrac{1}{3} + 3i \right)}^{3}}$.

Ans:

$\Rightarrow $${{\left( \dfrac{1}{3} + 3i \right)}^{3}} = {{\left( \dfrac{1}{3} \right)}^{3}} + {{\left( 3i \right)}^{3}} + 3\left( \dfrac{1}{3} \right)\left( 3i \right)\left( \dfrac{1}{3} + 3i \right) = \dfrac{1}{27} + 27{{i}^{3}} + 3i\left( \dfrac{1}{3} + 3i \right) = \dfrac{1}{27} + 27\left( - i \right) + i + 9{{i}^{2}} \left[ {{i}^{3}} = - i \right] = \dfrac{1}{27} - 27i + i + 9{{i}^{2}} \left[ {{i}^{2}} = - 1 \right] = \left( \dfrac{1}{27} - 9 \right) + i\left( - 27 + 1 \right) a + ib$$= \dfrac{- 242}{27} - 26i$

10. ${{\left( - 2 - \dfrac{1}{3}i \right)}^{3}}$.

Ans:

$\Rightarrow $${{\left( -2 - \dfrac{1}{3}i \right)}^{3}} = {{\left( - 1 \right)}^{3}}\left( 2 + \dfrac{ 1 }{3}i \right)^3 = - \left[ {{2}^{3}} + {{\left( \dfrac{i}{3} \right)}^{3}} + 3\left( 2 \right)\left( \dfrac{i}{3} \right)\left( 2 + \dfrac{i}{3} \right) \right] = - \left[ 8 + \dfrac{{{i}^{3}}}{27} + 2i\left( 2 +\dfrac{i}{3} \right) \right] \left[ {{i}^{3}} = - i \right] = - \left[ 8 - \dfrac{i}{27} + 4i - \dfrac{2{{i}^{2}}}{3} \right] \left[ {{i}^{2}} = - 1 \right] = - \left[ \dfrac{22}{3} + \dfrac{107i}{27} \right] a + ib$$= -\dfrac{22}{3} - \dfrac{107}{27}i$

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. $4 - 3i\text{ }.$

Ans:

Let $z = 4 - 3i\text{ }$

Then,

$\overline{z} = 4 + 3i$and $\left| z \right| = {{4}^{2}} + {{\left( - 3 \right)}^{2}} = 16 + 9 = 25$

The multiplicative inverse of $4 - 3i\text{ }$is given by

$\Rightarrow $${{z}^{- 1}} = \dfrac{\overline{z}}{{{\left| z \right|}^{2}}} $= \dfrac{4 + 3i}{25}$ $= \dfrac{4}{25} + \dfrac{3}{25}i$ 12. \sqrt{5} + 3i. Ans: Let z = \sqrt{5} + 3i Then, \overline{z} = \sqrt{5} - 3i \left| z \right| = {{\left( \sqrt{5} \right)}^{2}} +{{\left( 3 \right)}^{2}} = 5 + 9 = 14 The multiplicative inverse of the complex number \sqrt{5} + 3i is given by {{z}^{- 1}} = \dfrac{\overline{z}}{{{\left| z \right|}^{2}}} = \dfrac{\sqrt{5} - 3i}{14} = \dfrac{\sqrt{5}}{14} - \dfrac{3}{14}i 13. - i . Ans: Let z = - i Then, \overline{z} = i {{\left| z \right|}^{2}} = 1 The multiplicative inverse of the complex number - i \Rightarrow$${{z}^{- 1}} = \dfrac{\overline{z}}{{{\left| z \right|}^{2}}}$

$= \dfrac{i}{1}$

$= i$

14. Express the following expression in the form of $a + ib .$

$\dfrac{\left( 3 + i\sqrt{5} \right)\left( 3 - i\sqrt{5} \right)}{\left( \sqrt{3} + \sqrt{2 }i \right) - \left( \sqrt{3} - i \sqrt{2} \right)}$

Ans:

$\Rightarrow$$\dfrac{\left( 3 + i\sqrt{5} \right)\left( 3 - i\sqrt{5} \right)}{\left( \sqrt{3} + \sqrt{2 }i \right) - \left( \sqrt{3} - i \sqrt{2} \right)}$

$= \dfrac{\left( {{3}^{2}} - {{\left( i\sqrt{5} \right)}^{2}} \right)}{\left( \sqrt{3} + \sqrt{2 }i \right) - \left( \sqrt{3} - i \sqrt{2} \right)}$            $\left[ \left( a + b \right)\left( a - b \right) = {{a}^{2}} - {{b}^{2}} \right]$

$=\,\dfrac{\left( 9 - 5{{i}^{2}} \right)}{2\sqrt{2}i}$

$= \dfrac{9 - 5\left( - 1 \right)}{2\sqrt{2}i} \left[ {{i}^{2}} = - 1 \right]$

$= \dfrac{14i}{2\sqrt{2}{{i}^{2}}}$

$= \dfrac{14i}{2\sqrt{2}\left( - 1 \right)}$

$= \dfrac{- 7i}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}$

$= \dfrac{- 7\sqrt{2}i}{2}$

Conclusion

In conclusion, Class 11 Maths Chapter 4 Complex Numbers Exercise 4.1 has provided you with a comprehensive introduction to complex numbers. You have learned how to represent complex numbers in the form a+bi, perform basic operations such as addition, subtraction, multiplication, and division, and understand their graphical representation on the complex plane. Exercise 4.1 Class 11 Maths NCERT Solutions Chapter 4 will help students to gain foundational skills in handling complex numbers are crucial for solving quadratic equations and exploring more advanced topics in mathematics.

Class 11 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions Miscellaneous Exercise 14 Questions & Solutions

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations Exercise 4.1

1. Why should one opt for the NCERT solution for class 11 Maths chapter 4 complex number and quadratic equation exercise 4.1 by Vedantu?

When it comes to exam preparation, the NCERT solution for Class 11 Maths NCERT Solutions Chapter 4 Exercise 4.1 from Vedantu is thought to be the best choice of CBSE students. There are numerous exercises in this chapter. On this page in PDF format we have the exercise 4.1 class 11 math NCERT solutions. This solution is available for download at your convenience or you can assess it directly from the Vedanta website or app to study it.

2. Why should I practice Class 11 math NCERT chapter 4 complex number and quadratic equation exercise 4.1?

The NCERT book and solution have been prepared by the best and most highly skilled educational scholars and half created the content in such a way that all the math concepts could be understood by each student. Practicing exercise 4.1 will help you include your basics about the concept of complex numbers and quadratic equations and will help you go through the chapter. The NCERT math book has explained the concept of complex numbers on which the first exercise is based on with the help of solved examples that are written in the easiest way.

3. Where can I find the Solutions for Class 11 Maths NCERT Solutions Chapter 4 Exercise 4.1?

NCERT solution for Class 11 Chapter 4 Maths Exercise 4.1 Solutions complex number and quadratic equation are free to download in PDF format from vedantu. Teachers at vedantu are highly skilled and experts in their subjects and they have curated these NCERT solutions according to the latest CBSE pattern and guidelines.

4. How many complex numbers are there?

Hindi complex numbers are based on the nature of the real part and imaginary part any complex number can be classified into four types such as imaginary number, zero complex number, purely imaginary number, and purely real number.

5. Is 3 a complex number?

Yes, 3 is a complex number because real imaginary numbers combine to form complex numbers. That is why all imaginary numbers are also complex numbers. This also means that only real numbers are complex numbers for which intense 3 + 0i is equal to 3 but it is also a complex number because it has an imaginary part.

6. Why is it important to learn about complex numbers?

Complex numbers are important because they extend the concept of real numbers and are used in many areas of mathematics, physics, engineering, and other sciences. They are essential for solving certain types of equations and understanding advanced mathematical concepts.

7. How do you represent a complex number on the complex plane?

A complex number a+bi is represented on the complex plane by plotting the point (a,b), where a is the real part and b is the imaginary part. Understand more about complex plane in Class 11 Maths Chapter 4 Complex Numbers Exercise 4.1 .

8. What are the basic operations involving complex numbers covered in Exercise 4.1 Class 11 Maths NCERT Solutions Chapter 4?

The basic operations involving complex numbers include addition, subtraction, multiplication, and division.

9. How do you add two complex numbers?

To add two complex numbers a+bii and c+di, add the real parts and the imaginary parts separately: (a+c)+(b+d)i.

10. How do you subtract one complex number from another?

To subtract c+di from a+bi, subtract the real parts and the imaginary parts separately: (a−c)+(b−d)i.