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NCERT Solutions

Class 11

Maths

Chapter 4 Complex Numbers And Quadratic Equations Exercise 4.1

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations Ex 4.1

Q: 5. Why is every real number a complex number, but not every complex number is a real number?

This is because the set of complex numbers is a superset of real numbers. A complex number is in the form a + ib. A real number, like 5, can be written as 5 + 0i, where its imaginary part 'b' is zero. This fits the definition of a complex number. However, a complex number like 2 + 3i has a non-zero imaginary part, so it cannot be expressed as just a real number.

Q: 6. In the NCERT Solutions for Class 11 Maths Chapter 4, why is it necessary to use the Argand plane to represent complex numbers?

The Argand plane, or complex plane, is necessary because a complex number has two components: a real part (a) and an imaginary part (b). A simple one-dimensional number line can only represent real numbers. The Argand plane uses a two-dimensional Cartesian coordinate system where the x-axis represents the real part and the y-axis represents the imaginary part. This allows for a unique geometric visualisation of every complex number as a point (a, b).

Q: 7. What is the specific role of the conjugate when dividing two complex numbers?

The conjugate of a complex number (z = a + ib) is (z̄ = a - ib). Its primary role in division is to eliminate the imaginary part 'i' from the denominator. When you multiply a complex number by its conjugate, the result is always a real number: (a + ib)(a - ib) = a² + b². This process transforms the denominator into a real number, making it possible to express the final result in the standard a + ib form.

Q: 8. What are the main topics covered in the NCERT Solutions for Class 11 Maths Chapter 4?

The NCERT Solutions for Class 11 Maths Chapter 4, "Complex Numbers and Quadratic Equations," cover several key topics as per the 2025-26 CBSE syllabus. These include:The concept of the imaginary unit 'i' and its integral powers.Algebra of complex numbers: addition, subtraction, multiplication, and division.Finding the modulus and conjugate of a complex number.The Argand plane and polar representation of complex numbers.Solving quadratic equations where the discriminant is negative, leading to complex roots.

Q: 9. Can the square root of a negative number be a real number? How does Chapter 4 address this?

No, the square root of a negative number cannot be a real number. The set of real numbers does not have a solution for expressions like √-1. Chapter 4 introduces the concept of imaginary numbers to solve this exact problem. It defines the imaginary unit 'i' as the solution to √-1. This extends the number system to complex numbers, allowing us to find solutions for any quadratic equation, regardless of the sign of its discriminant.

Q: 10. How do you apply the algebraic properties of complex numbers when solving problems in the NCERT exercises?

The NCERT exercises require applying several algebraic properties to simplify expressions. For instance:The commutative property (z₁ + z₂ = z₂ + z₁) and associative property (z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃) for addition and multiplication allow you to regroup and reorder terms for easier calculation.The distributive property (z₁(z₂ + z₃) = z₁z₂ + z₁z₃) is fundamental for expanding expressions involving multiplication of complex numbers.Understanding the properties of the multiplicative identity (1) and additive identity (0) helps in simplifying complex equations step-by-step as shown in the solutions.

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

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)

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

CBSE Class 11 Maths Chapter 4 Other Study Materials

S. No	Important Links for Chapter 4 Complex Numbers and Quadratic Equations
1	Class 11 Complex Numbers and Quadratic Equations Important Questions
2	Class 11 Complex Numbers and Quadratic Equations Revision Notes
3	Class 11 Complex Numbers and Quadratic Equations Important Formulas
4	Class 11 Complex Numbers and Quadratic Equations NCERT Exemplar Solution
5	Class 11 Complex Numbers and Quadratic Equations RD Sharma Solutions
6	Class 11 Complex Numbers and Quadratic Equations RS Aggarwal 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.

S. No	NCERT Solutions Class 11 Maths All Chapters
1	Chapter 1 - Sets Solutions
2	Chapter 2 - Relations and Functions Solutions
3	Chapter 3 - Trigonometric Functions Solutions
4	Chapter 4 - Complex Numbers and Quadratic Equations Solutions
5	Chapter 5 - Linear Inequalities Solutions
6	Chapter 6 - Permutations and Combinations Solutions
7	Chapter 7 - Binomial Theorem Solutions
8	Chapter 8 - Sequences and Series Solutions
9	Chapter 9 - Straight Lines Solutions
10	Chapter 10 - Conic Sections Solutions
11	Chapter 11 - Introduction to Three Dimensional Geometry Solutions
12	Chapter 12 - Limits and Derivatives Solutions
13	Chapter 13 - Statistics Solutions
14	Chapter 14 - Probability Solutions

Important Related Links for CBSE Class 11 Maths

S.No.	Important Study Material for Maths Class 11
1	CBSE Class 11 Maths NCERT Books
2	CBSE Class 11 Maths Revision Notes
3	CBSE Class 11 Maths Important Questions
4	CBSE Class 11 Maths NCERT Solutions
5	CBSE Class 11 Maths NCERT Exemplar Solution
6	CBSE Class 11 Maths RS Aggarwal Solutions
7	CBSE Class 11 Maths RD Sharma Solutions
8	CBSE Class 11 Maths Formulas
9	CBSE Class 11 Maths Syllabus
10	CBSE Class 11 Maths Sample Papers

FAQs on NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations Ex 4.1

1. What is the standard form of a complex number as per the NCERT textbook for Class 11 Maths?

As per the NCERT Class 11 Maths textbook for the 2025-26 session, any complex number, denoted by z, is expressed in the standard form a + ib. In this form, 'a' is the real part, denoted as Re(z), and 'b' is the imaginary part, denoted as Im(z). The term 'i' represents the imaginary unit, which is equal to the square root of -1 (√-1).

2. How do you find the multiplicative inverse of a complex number in NCERT Solutions for Chapter 4?

To find the multiplicative inverse of a non-zero complex number z = a + ib, you use the formula: z⁻¹ = a/(a² + b²) + i(-b/(a² + b²)). This is derived by multiplying the numerator and denominator of 1/(a + ib) by its conjugate (a - ib). The NCERT solutions for Chapter 4 provide step-by-step examples of this correct method.

3. What is the correct method to express a power of 'i', such as i⁻³⁹, in the standard form a + ib?

The correct method involves using the cyclical nature of powers of i (i¹=i, i²=-1, i³=-i, i⁴=1). The steps are as follows:

First, make the exponent positive: i⁻³⁹ = 1/i³⁹.
Next, express the exponent in terms of a multiple of 4: i³⁹ = i^(4*9 + 3) = (i⁴)⁹ * i³.
Since i⁴ = 1, this simplifies to 1⁹ * i³ = i³.
We know i³ = -i. So, 1/i³⁹ becomes 1/(-i).
To simplify, multiply the numerator and denominator by i: (1 * i) / (-i * i) = i / (-i²) = i / (-(-1)) = i.
The final standard form is 0 + 1i.

4. How do you solve a quadratic equation that has complex roots, according to the CBSE pattern?

For a quadratic equation of the form ax² + bx + c = 0, you first calculate the discriminant (D), where D = b² - 4ac. If D < 0, the roots are complex. The solutions are found using the quadratic formula: x = [-b ± √D] / 2a. Since D is negative, we can write √D as √( |D| * -1) = i√|D|. The two complex roots will be [-b + i√|D|] / 2a and [-b - i√|D|] / 2a.

5. Why is every real number a complex number, but not every complex number is a real number?

This is because the set of complex numbers is a superset of real numbers. A complex number is in the form a + ib. A real number, like 5, can be written as 5 + 0i, where its imaginary part 'b' is zero. This fits the definition of a complex number. However, a complex number like 2 + 3i has a non-zero imaginary part, so it cannot be expressed as just a real number.

6. In the NCERT Solutions for Class 11 Maths Chapter 4, why is it necessary to use the Argand plane to represent complex numbers?

The Argand plane, or complex plane, is necessary because a complex number has two components: a real part (a) and an imaginary part (b). A simple one-dimensional number line can only represent real numbers. The Argand plane uses a two-dimensional Cartesian coordinate system where the x-axis represents the real part and the y-axis represents the imaginary part. This allows for a unique geometric visualisation of every complex number as a point (a, b).

7. What is the specific role of the conjugate when dividing two complex numbers?

The conjugate of a complex number (z = a + ib) is (z̄ = a - ib). Its primary role in division is to eliminate the imaginary part 'i' from the denominator. When you multiply a complex number by its conjugate, the result is always a real number: (a + ib)(a - ib) = a² + b². This process transforms the denominator into a real number, making it possible to express the final result in the standard a + ib form.

8. What are the main topics covered in the NCERT Solutions for Class 11 Maths Chapter 4?

The NCERT Solutions for Class 11 Maths Chapter 4, "Complex Numbers and Quadratic Equations," cover several key topics as per the 2025-26 CBSE syllabus. These include:

The concept of the imaginary unit 'i' and its integral powers.
Algebra of complex numbers: addition, subtraction, multiplication, and division.
Finding the modulus and conjugate of a complex number.
The Argand plane and polar representation of complex numbers.
Solving quadratic equations where the discriminant is negative, leading to complex roots.

9. Can the square root of a negative number be a real number? How does Chapter 4 address this?

No, the square root of a negative number cannot be a real number. The set of real numbers does not have a solution for expressions like √-1. Chapter 4 introduces the concept of imaginary numbers to solve this exact problem. It defines the imaginary unit 'i' as the solution to √-1. This extends the number system to complex numbers, allowing us to find solutions for any quadratic equation, regardless of the sign of its discriminant.

10. How do you apply the algebraic properties of complex numbers when solving problems in the NCERT exercises?

The NCERT exercises require applying several algebraic properties to simplify expressions. For instance:

The commutative property (z₁ + z₂ = z₂ + z₁) and associative property (z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃) for addition and multiplication allow you to regroup and reorder terms for easier calculation.
The distributive property (z₁(z₂ + z₃) = z₁z₂ + z₁z₃) is fundamental for expanding expressions involving multiplication of complex numbers.
Understanding the properties of the multiplicative identity (1) and additive identity (0) helps in simplifying complex equations step-by-step as shown in the solutions.