NCERT Solutions for Class 11 Maths Chapter 5
Exercise 5.1 Page No: 103
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
1. (5i) (-3/5i)
Solution:
(5i) (-3/5i) = 5 x (-3/5) x i2
= -3 x -1 [i2 = -1]
= 3
Hence,
(5i) (-3/5i) = 3 + i0
2. i9 + i19
Solution:
i9 + i19 = (i2)4. i + (i2)9. i
= (-1)4 . i + (-1)9 .i
= 1 x i + -1 x i
= i – i
= 0
Hence,
i9 + i19 = 0 + i0
3. i-39
Solution:
i-39 = 1/ i39 = 1/ i4 x 9 + 3 = 1/ (19 x i3) = 1/ i3 = 1/ (-i) [i4 = 1, i3 = -I and i2 = -1]
Now, multiplying the numerator and denominator by i we get
i-39 = 1 x i / (-i x i)
= i/ 1 = i
Hence,
i-39 = 0 + i
4. 3(7 + i7) + i(7 + i7)
Solution:
3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i2 7
= 21 + i28 – 7 [i2 = -1]
= 14 + i28
Hence,
3(7 + i7) + i(7 + i7) = 14 + i28
5. (1 – i) – (–1 + i6)
Solution:
(1 – i) – (–1 + i6) = 1 – i + 1 – i6
= 2 – i7
Hence,
(1 – i) – (–1 + i6) = 2 – i7
6.
Solution:
7. 
Solution:
8. (1 – i)4
Solution:
(1 – i)4 = [(1 – i)2]2
= [1 + i2 – 2i]2
= [1 – 1 – 2i]2 [i2 = -1]
= (-2i)2
= 4(-1)
= -4
Hence, (1 – i)4 = -4 + 0i
9. (1/3 + 3i)3
Solution:
Hence, (1/3 + 3i)3 = -242/27 – 26i
10. (-2 – 1/3i)3
Solution:
Hence,
(-2 – 1/3i)3 = -22/3 – 107/27i
Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.
11. 4 – 3i
Solution:
Let’s consider z = 4 – 3i
Then,
= 4 + 3i and
|z|2 = 42 + (-3)2 = 16 + 9 = 25
Thus, the multiplicative inverse of 4 – 3i is given by z-1
12. √5 + 3i
Solution:
Let’s consider z = √5 + 3i
|z|2 = (√5)2 + 32 = 5 + 9 = 14
Thus, the multiplicative inverse of √5 + 3i is given by z-1
13. – i
Solution:
Let’s consider z = –i
Thus, the multiplicative inverse of –i is given by z-1
14. Express the following expression in the form of a + ib:
Solution:
Exercise 5.2 Page No: 108
Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
1. z = – 1 – i √3
Solution:
2. z = -√3 + i
Solution:
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
3. 1 – i
Solution:
4. – 1 + i
Solution:
5. – 1 – i
Solution:
6. – 3
Solution:
7. 3 + i
Solution:
8. i
Solution:
Exercise 5.3 Page No: 109
Solve each of the following equations:
1. x2 + 3 = 0
Solution:
Given quadratic equation,
x2 + 3 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 0, and c = 3
So, the discriminant of the given equation will be
D = b2 – 4ac = 02 – 4 × 1 × 3 = –12
Hence, the required solutions are:
2. 2x2 + x + 1 = 0
Solution:
Given quadratic equation,
2x2 + x + 1 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 2, b = 1, and c = 1
So, the discriminant of the given equation will be
D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7
Hence, the required solutions are:
3. x2 + 3x + 9 = 0
Solution:
Given quadratic equation,
x2 + 3x + 9 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 3, and c = 9
So, the discriminant of the given equation will be
D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27
Hence, the required solutions are:
4. –x2 + x – 2 = 0
Solution:
Given quadratic equation,
–x2 + x – 2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = –1, b = 1, and c = –2
So, the discriminant of the given equation will be
D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7
Hence, the required solutions are:
5. x2 + 3x + 5 = 0
Solution:
Given quadratic equation,
x2 + 3x + 5 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = 3, and c = 5
So, the discriminant of the given equation will be
D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11
Hence, the required solutions are:
6. x2 – x + 2 = 0
Solution:
Given quadratic equation,
x2 – x + 2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = 1, b = –1, and c = 2
So, the discriminant of the given equation is
D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7
Hence, the required solutions are
7. √2x2 + x + √2 = 0
Solution:
Given quadratic equation,
√2x2 + x + √2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = 1, and c = √2
So, the discriminant of the given equation is
D = b2 – 4ac = (1)2 – 4 × √2 × √2 = 1 – 8 = –7
Hence, the required solutions are:
8. √3x2 – √2x + 3√3 = 0
Solution:
Given quadratic equation,
√3x2 – √2x + 3√3 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √3, b = -√2, and c = 3√3
So, the discriminant of the given equation is
D = b2 – 4ac = (-√2)2 – 4 × √3 × 3√3 = 2 – 36 = –34
Hence, the required solutions are:
9. x2 + x + 1/√2 = 0
Solution:
Given quadratic equation,
x2 + x + 1/√2 = 0
It can be rewritten as,
√2x2 + √2x + 1 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = √2, and c = 1
So, the discriminant of the given equation is
D = b2 – 4ac = (√2)2 – 4 × √2 × 1 = 2 – 4√2 = 2(1 – 2√2)
Hence, the required solutions are:
10. x2 + x/√2 + 1 = 0
Solution:
Given quadratic equation,
x2 + x/√2 + 1 = 0
It can be rewritten as,
√2x2 + x + √2 = 0
On comparing it with ax2 + bx + c = 0, we have
a = √2, b = 1, and c = √2
So, the discriminant of the given equation is
D = b2 – 4ac = (1)2 – 4 × √2 × √2 = 1 – 8 = -7
Hence, the required solutions are:
Miscellaneous Exercise Page No: 112
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 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:
Thus, the least positive integer is 1.
Therefore, the least positive integral value of m is 4 (= 4 × 1).