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

Chapter 5 Complex Numbers and Quadratic Equations of Class 11 Maths is categorized under the CBSE Syllabus for 2022-23. 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

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.

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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 zRe 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 equation 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 = 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 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 = 2x

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).