NCERT Solutions Class 11th Maths Chapter – 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise
| Textbook | NCERT |
| class | Class – 11th |
| Subject | Mathematics |
| Chapter | Chapter – 5 |
| Chapter Name | Complex Numbers and Quadratic Equations |
| grade | Class 11th Maths solution |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 11th Maths Chapter – 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise
?Chapter – 5?
✍Complex Numbers and Quadratic Equations✍
?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:
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 + 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 = 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:
