NCERT Solutions Class 11 Maths Chapter 4 Complex Numbers & Quadratic Equations
NCERT Solutions for Class 11 are highly important for making students understand the tough concepts of Maths. One of the most important chapters in class 11 maths chapter 4 is Complex Numbers and Quadratic Equations. This chapter forms the basis of higher mathematical studies. Sets, complex numbers, and quadratic equations constitute some of the most important concepts for examinations. These NCERT solutions in-built with detailed explanations and step-by-step problem-solving methods so that all students can face any question confidently. Moreover, to make it more accessible, one can download the class 11 maths chapter 4 PDF to study anywhere. Orchids International School lays emphasis on strong conceptual learning; therefore, these NCERT solutions may prove to be of immense value to all of them in achieving greatness in sets, complex numbers, and quadratic equations, and many more, other than building a great academic base.
Access Answers to NCERT Solutions Class 11 Maths Chapter 4 Complex Numbers & Quadratic Equations
Students can access the NCERT Solutions Class 11 Maths Chapter 4 Complex Numbers & Quadratic Equations. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Complex Numbers & Quadratic Equations
i9 + i19
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(7 + i7) + i(7 + i7)
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
(1 – i) – (–1 + i6)
(1 – i) – (–1 + i6) = 1 – i + 1 – i6
= 2 – i7
Hence,
(1 – i) – (–1 + i6) = 2 – i7
Express each of the complex numbers given in Exercises 1 to 10 in the form a + ib.
1. (5i) (-3/5i)
(5i) (-3/5i) = 5 x (-3/5) x i2
= -3 x -1 [i2 = -1]
= 3
Hence,
(5i) (-3/5i) = 3 + i0
i-39
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
(1 – i)4
(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
(1/3 + 3i)3
Hence, (1/3 + 3i)3 = -242/27 – 26i
(-2 – 1/3i)3
Hence,
(-2 – 1/3i)3 = -22/3 – 107/27i
Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.
11. 4 – 3i
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
√5 + 3i
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
– i
Let’s consider z = –i
Thus, the multiplicative inverse of –i is given by z-1
Express the following expression in the form of a + ib:
Exercise 5.2
Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
1. z = – 1 – i √3
z = -√3 + i
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
3. 1 – i
4. – 1 + i
5. – 1 – i
6. – 3
7. 3 + i
8. i
Exercise 5.3
Solve each of the following equations:
1. x2 + 3 = 0
Given the 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
2x2 + x + 1 = 0
Given the 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
x2 + 3x + 9 = 0
Given the 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
–x2 + x – 2 = 0
Given the 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
x2 + 3x + 5 = 0
Given the 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
x2 – x + 2 = 0
Given the 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
√2x2 + x + √2 = 0
Given the 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
√3x2 – √2x + 3√3 = 0
Given the 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
x2 + x + 1/√2 = 0
Given the 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
x2 + x/√2 + 1 = 0
Given the 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
For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Convert the following into the polar form:
(i) 
Solve each of the equations in Exercises 6 to 9.
3x2 – 4x + 20/3 = 0
Given the 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
21x2 – 28x + 10 = 0
Given the 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
x2 – 2x + 3/2 = 0
Given the 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
27x2 – 10x + 1 = 0
Given the 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
If z1 = 2 – i, z2 = 1 + i, find
Given, z1 = 2 – i, z2 = 1 + i
Let z1 = 2 – i, z2 = -2 + i. Find
(i) 
Find the modulus and argument of the complex number.
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.
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.
Find the modulus of
If (x + iy)3 = u + iv, then show that
If α and β are different complex numbers with |β| = 1, then find
Find the number of non-zero integral solutions of the equation |1 – i|x = 2x.
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.
If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
If, then find the least positive integral value of m.
Thus, the least positive integer is 1.
Therefore, the least positive integral value of m is 4 (= 4 × 1).
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