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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

Chapter 5 Complex Numbers and Quadratic Equations of Class 11 Maths is prescribed for study in the CBSE Syllabus for 2023-24. The textbook of Class 11 Maths contains an exercise at the end of each chapter. This exercise contains questions that cover all the topics in the chapter. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations is based on the following topics:

1. Complex Numbers
2. Algebra of Complex Numbers
3. The Modulus and the Conjugate of a Complex Number
4. Argand Plane and Polar Representation
5. Quadratic Equations

Mathematics is a subject that needs more and more practice to understand the problem-solving method. Practising with the help of NCERT Solutions for Class 11 Maths will help the students in understanding the concepts and problems better, in turn, helping them to score high marks in the board exams.

NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise

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Access other exercise solutions of Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations

Practice more problems related to the concepts of complex numbers and quadratic equations using the NCERT Class 11 Solutions for Maths links given here.

Exercise 5.1 Solutions 14 Questions

Exercise 5.2 Solutions 8 Questions

Exercise 5.3 Solutions 10 Questions

Access Solutions for Class 11 Maths Chapter 5 Miscellaneous Exercise

1.

Solution:

2. For any two complex numbers z₁ and z₂, prove that

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Solution:

3. Reduce to the standard form

Solution:

4.

Solution:

5. Convert the following in the polar form:

(i) , (ii) 

Solution:

Solve each of the equations in Exercises 6 to 9.

6. 3x² – 4x + 20/3 = 0

Solution:

Given quadratic equation, 3x² – 4x + 20/3 = 0

It can be re-written as: 9x² – 12x + 20 = 0

On comparing it with ax² + bx + c = 0, we get

a = 9, b = –12, and c = 20

So, the discriminant of the given equation will be

D = b² – 4ac = (–12)² – 4 × 9 × 20 = 144 – 720 = –576

Hence, the required solutions are

7. x² – 2x + 3/2 = 0

Solution:

Given quadratic equation, x² – 2x + 3/2 = 0

It can be re-written as 2x² – 4x + 3 = 0

On comparing it with ax² + bx + c = 0, we get

a = 2, b = –4, and c = 3

So, the discriminant of the given equation will be

D = b² – 4ac = (–4)² – 4 × 2 × 3 = 16 – 24 = –8

Hence, the required solutions are

8. 27x² – 10x + 1 = 0

Solution:

Given quadratic equation, 27x² – 10x + 1 = 0

On comparing it with ax² + bx + c = 0, we get

a = 27, b = –10, and c = 1

So, the discriminant of the given equation will be

D = b² – 4ac = (–10)² – 4 × 27 × 1 = 100 – 108 = –8

Hence, the required solutions are

9. 21x² – 28x + 10 = 0

Solution:

Given quadratic equation, 21x² – 28x + 10 = 0

On comparing it with ax² + bx + c = 0, we have

a = 21, b = –28, and c = 10

So, the discriminant of the given equation will be

D = b² – 4ac = (–28)² – 4 × 21 × 10 = 784 – 840 = –56

Hence, the required solutions are

10. If z₁ = 2 – i, z₂ = 1 + i, find

Solution:

Given, z₁ = 2 – i, z₂ = 1 + i

11.

Solution:

12. Let z₁ = 2 – i, z₂ = -2 + i. Find

(i) , (ii) 

Solution:

13. Find the modulus and argument of the complex number

Solution:

14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.

Solution:

Let’s assume z = (x – iy) (3 + 5i)

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of x and y are 3 and –3 respectively.

15. Find the modulus of

Solution:

16. If (x + iy)³ = u + iv, then show that

Solution:

17. If α and β are different complex numbers with |β| = 1, then find

Solution:

18. Find the number of non-zero integral solutions of the equation |1 – i|^x = 2^x

Solution:

Therefore, 0 is the only integral solution of the given equation.

Hence, the number of non-zero integral solutions of the given equation is 0.

19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

Solution:

20. If, then find the least positive integral value of m.

Solution:

Thus, the least positive integer is 1.

Therefore, the least positive integral value of m is 4 (= 4 × 1).