# NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations

## Chapter 5 Complex Numbers and Quadratic Equations Class 11 Mathematics NCERT Solutions

Class 11 Mathematics students should refer to the following NCERT questions with answers for Chapter 5 Complex Numbers and Quadratic Equations in Class 11. These NCERT Solutions with answers for Class 11 Mathematics will come in exams and help you to score good marks

### Chapter 5 Complex Numbers and Quadratic Equations NCERT Solutions Class 11 Mathematics

Question. Express the given complex number in the form a + ib: (5i) (-3/5)i
(5i) (-3/5)i = -5 × (3/5) × i × i
= -3i2
= -3(-1)  [i2 = -1]
= 3

Question. Express the given complex number in the form a + ib : i9 + i19
i9 + i19 = i4×2+1 + i4×4+3
= (i4)2i + (i4)4i3
= 1×i + 1×(-i)  [i4 = 1, i3 = -i]
= i + (-i)
= 0

Question. Express the given complex number in the form a +ib : i-39

Question. Express the given complex number in the form a + ib : 3(7 + i7) + i(7 + i7)
3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i2
= 21 + 28i + 7 × (-1)  [∵i2  = -1]
= 14 + 28i

Question. Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)
(1 - i) - (-1 + i6) = 1 - i + 1 - 6i
= 2 - 7i

Question. Express the given complex number in the form a + ib: (1/5 +i 2/5) - (4 + i 5/2)

Question. Express the given complex number in the form a + ib: [(1/3 + 7/3) + (4 + i 1/3)] - (-4/3 + i

8. Express the given complex number in the form a + ib : (1 - )4

(1 -i)2 = [(1 - i)2]2

= [12 + i2 + 2i2]2
= [1 - 1 - 2i]2
= (-2i)2
= (-2i)×(-2i)
= 4i2 = -4  [i2  = -1]

Question. Express the given complex number in the form a + ib : (1/3 + 3i)3

Question. Express the given complex number in the form a + ib : [-2 - (1/3)i]3

Question. Find the multiplicative inverse of the complex number 4 - 3i
Let z = 4 - 3i
Then, z  = 4 + 3i and |z|2 = 42 + (-3)2 = 16 + 9 = 25
Therefore, the multiplicative inverse of 4 - 3i is given by

Question. Find the multiplicative inverse of the complex number √5 + 3i
Let z = √5 + 3i
Then, z = √5 + 3i and |z|2 = (√5)2 + 32 = 5 + 9= 14
Therefore, the multiplicative inverse of √5 + 3i is given by

Question. Find the multiplicative inverse of the complex number –i
Let z = -i
Then, z  = i and |z|2 = 12 = 1
Therefore, the multiplicative inverse of -i is given by
z-1 = z /|z|2 = i/1 = i

Question. Express the following expression in the form of a + ib.
[(3 + i√5)(3 - i√5)]/[(√3 + √2i)- (√3 - i√2)]

Exercise 5.2

Question. Find the modulus and the argument of the complex number z = -1 - i√3
z = -1- i√3
Let r cos θ = -1 and r sin θ = -√3
On squaring and adding, we obtain
(r cos θ)2 + (r sin θ)2 = (-1)2 + (-√3)2
⇒ r2 (cos2 θ + sin2 θ) = 1 + 3
⇒  r2 = 4  [cos2 θ + sin2 θ = 1]
⇒ r = √4 = 2  [Conventionally, r > 0]
∴ Modulus = 2
∴ 2 cos θ = -1 and 2sin θ = - √3
⇒ cos θ = -1/2  and sin θ = -√3/2
Since both the values of sin θ and cos θ are negative and sin θ and cos θ are negative in III quadrant,
Argument = -(π - π/3) = -2π/3
Thus, the modulus and argument of the complex number -1 - √3 i are 2 and -2π/3 respectively.

Question. Find the modulus and the argument of the complex number z = -√3 + i
z = -√3 + i
Let r cos θ = -√3 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = (-√3)2 + 12
⇒ r2 = 3 + 1 = 4  [cos2 θ + sin2 θ = 1]
⇒ r = √4 = 2  [Conventionally, r > 0]
∴ Modulus = 2
∴ 2 cos θ = -√3 and 2sin θ = 1
⇒ cosθ = -√3/2 and sinθ = 1/2
∴ θ = π - π/6 = 5π/6  [As θ lies in the II quadrant]
Thus, the modulus and argument of the complex number -√3 + i are 2 and 5π/6 respectively.

Question. Convert the given complex number in polar form : 1 - i
1 - i
Let r cos θ = 1 and r sin θ = -1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = 12 + (-1)2
⇒ r2 (cos2 θ + sin2 θ) = 1 + 1
⇒ r2 = 2
⇒ r = √2  [Conventionally, r > 0]
∴ √2 cos θ = 1 and √2 sin θ = -1
⇒ cos θ = 1/√2  and sinθ = -1/√2
∴ θ = -π/4  [As θ lies in the IV quadrant]
∴ 1 - i = r cos θ + i r sinθ = √2 cos (-π/4)  + i√2sin(-π/4) = √2[cos (-π/4) + i sin(-π/4) ] This is the required polar form.

Question. Convert the given complex number in polar form : -1 + i
-1 + i
Let r cos θ = -1 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = (-1)2 + 12
⇒ r2 (cos2 θ + sin2 θ) = 1 + 1
⇒ r2 = 2
⇒ r = √2  [Conventionally, r > 0]
∴ √2 cos θ = -1 and √2sin θ = 1
⇒ cos θ = -1/√2 and sinθ = 1/√2
∴ θ = π - π/4 = 3π/4  [As θ lies in the II quadrant]
It can be written,
∴ -1 + i = r cos θ + i r sinθ

Question. Convert the given complex number in polar form : -1 - i .
-1-i
Let r cos θ = -1 and r sin θ = -1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = (-1)2  + (-1)2
⇒ r2 (cos2 θ + sin2 θ) = 1 + 1
⇒ r2 = 2
⇒ r = √2  [Conventionally, r > 0]
∴ √2 cos θ = -1 and √2sin θ = -1
⇒ cos θ = -1/√2 and sinθ = -1/√2
∴ θ = -(π - π/4) = -3π/4  [As θ lies in the III quadrant]
∴ -1-i = r cos θ + i r sin θ = √2 cos (-3π/4) + i√2 sin(-3π/4) = √2[cos(-3π/4) + i sin(-3π/4)]
This is the required polar form.

Question. Convert the given complex number in polar form: -3
-3
Let r cos c = -3 and r sin θ = 0
On squaring and adding we obtain
r2 cos2 θ + r2 sin2 θ = (-3)2
⇒ r2 (cos2 θ + sin2 θ) = 9
⇒ r2 = 9
⇒ r = √9 = 3  [Conventionally, r > 0]
∴ 3 cosθ = -3 and 3 sinθ = 0
⇒ cos θ= -1 and sinθ = 0
∴ θ = π
∴ -3 = r cos θ + i r sin θ = 3 cos π + B sinπ = 3(cos π + isin π)
This is the required polar form.

Question. Convert the given complex number in polar form : √3 + i
√3 + i
let r cos θ = √3 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = (√3)2 + 12
⇒ r2 (cos2 θ + sin2 θ) = 3 + 1
⇒ r2 = 4
⇒ r = √4 = 2   [Conventionally, r > 0]
∴ 2 cosθ = √3 and 2sinθ = 1
⇒ cos θ = √3/2 and sinθ = 1/2
∴ θ = π/6  [As θ lies n the I quadrant]
∴ √3 + i = r cos θ + i r sinθ = 2 cos (π/6 ) + i 2 sin (π/6) = 2[cos(π/6) + i sin(π/6)]
This is the required polar form.

Question. Convert the given complex number in polar form : i
i
Let r cosθ = 0 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = 02 + 12
⇒ r2 (cos2 θ + sin2 θ) = 1
⇒ r2 = 1
⇒ r = √1 = 1  [Conventionally, r > 0]
∴ cosθ = 0 and sinθ = 1
∴ θ = π/2
∴ i = r cosθ + i r sin θ = cos (π/2) + i sin (π/2)
This is the required polar form.

Exercise 5.3

Question. Solve the equation x2 + 3 = 0
The given quadratic equation is x2 + 3 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = 0, and c = 3
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 02 – 4 × 1 × 3 = –12
Therefore, the required solutions are

Question. Solve the equation 2x2 + x + 1 = 0
The given quadratic equation is 2x2 + x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 2, b = 1, and c = 1
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7
Therefore, the required solutions are

Question. Solve the equation x2 + 3x + 9 = 0
The given quadratic equation is x2 + 3x + 9 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = 3, and c = 9
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27
Therefore, the required solutions are

Question. Solve the equation –x2 + x – 2 = 0
The given quadratic equation is –x2 + – 2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = –1, b = 1, and c = –2
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7
Therefore, the required solutions are

Question. Solve the equation x2 + 3x + 5 = 0
The given quadratic equation is x2 + 3x + 5 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = 3, and c = 5
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11
Therefore, the required solutions are

6. Solve the equation x2 – x + 2 = 0
The given quadratic equation is x2 – x + 2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = –1, and c = 2
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7
Therefore, the required solutions are

Question. Solve the equation √2x2 + x + √2 = 0
The given quadratic equation is √2x2 + x + √2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = √2, b = 1, and c = √2
Therefore, the discriminant of the given equation is
D = b2 - 4ac = 12 - 4 × √2 × √2 = 1 - 8 =-7
Therefore, the required solution are

Question. Solve the equation √3x2 - √2x + 3√3 = 0
The given quadratic equation is √3x2 - √2x + 3√3 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = √3, b = -√2, and c = 3√3
Therefore, the discriminant of the given equation is
d = b2 - 4ac = (-√2)2 - 4(√3)(3√3) = 2 - 36 = -34
Therefore, the required solution are

Question. Solve the equation x2 + x + 1/√2 = 0
The given quadratic equation is x2 + x+ 1/√2 = 0
This equation can also be written as √2x2 + √2x+ 1 = 0
On comparing this equation  with ax2 + bx + c = 0, we obtain
a = √2, b = √2, and c = 1
∴ Discriminant (D) = b2 - 4ac = (√2)2 - 4×√2 ×1 = 2 - 4√2
Therefore, the required solutions are

Question. Solve the equation x2 + x/√2 + 1 = 0
The given quadratic equation is x2 + x/√2 + 1 = 0
This equation can also be written as √2x2 + x + √2 = 0
On comparing this equation with ax2 +bx + c = 0, we obtain
a = √2, b = 1, and c = √2
∴ Discriminant (D) = b2 - 4ac = 12 - 4×√2×√2 = 1 - 8 = -7
Therefore, the required solutions are

Miscellaneous Solutions

Question. Evaluate : [i18 + (1/i)25 ]3 .

= [-1 - i]3
= (-1)3 [1 +i]3
= -[13 + i3 + 3.1.i(1 + i)]
= -[1 + i3 + 3i + 3i2 ]
= -[1 - i + 3i -3]
= -[-2+2i]
= 2 -2i

Question. For any two complex numbersz1 and z2, prove that
Re (z1z2)= Re zRe z2 – Im z1 Im z2

Let z1 = x1 + iy1 and z2 = x2 + iy2
∴ z1z2 = (x1 + iy1)(x2 + iy2)
= x1 (x2 + iy2) + iy1 (x2 + iy2
= x1x2 + ix1y2 + iy1x2 + i2 y1y2
= x1x2 + ix1y2 + iy1x2 + y1y2 [i2 = -1]
= (x1x2 - y1y2) + i(x1y2 + y1x2
⇒ Re (z1z2) = x1x2 - y1x2
⇒ Re (z1z2) = Re z1 Re z2 - Im z1 Im z2
Hence, proved.

Question. Reduce [(1/(1-4i) - 2/(1 + i)][(3 - 4i)/(5 +i)] to the standard form.

Question. If x - iy = √(a - ib)/(c-id) prove that (x2 + y2)2  = (a2 + b2)/(c2 + d2

Question. Convert the following in the polar form :
(i) (1 + 7i)/(2 - i)2 ,
(ii) (1 + 3i)/(1 - 2i)

Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒ r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2  [cos2 θ + sin2 θ = 1]
⇒ r = √2  [Conventionally, r > 0]
∴ √2 cosθ = -1 and √2 sinθ = 1
⇒ cos θ = -1/√2 and sin θ = 1/√2
∴ θ = π - π/4 = 3π/4   [As θ lies in II quadrant]
∴ z = r cos θ + i r sin θ
= √2 cos (3π/4) + i √2sin (3π/4) = √2[ cos(3π/4) + i sin(3π/4)
This is the required polar form.
(ii) Here, z = (1 + 3i)/(1 - 2i)

Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
⇒ r = √2   [Conventionally, r > 0]
∴ √2 cosθ = -1 and √2sinθ = 1
⇒ cos θ = -1/√2 and sinθ = 1/√2
∴ θ = π - π4 = 3π/4  [As θ lies in II quadrant]
∴ z = r cos θ + i r sin θ
= √2 cos 3π/4 + i√2 sin3π/4 = √2 [cos (3π/4 + i sin(3π/4)]
This is the required polar form.

Question. Solve the equation 3x2 - 4x + 20/3 = 0
The given quadratic equation is 3x2 - 4x + 20/3 = 0
This equation can also be written as 9x2 - 12x + 20 = 0
On comparing this equation with ax2 + bx + c = 0 , we obtain
a = 9 b = -12 and c = 20
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-12)2 - 4× 9 × 20 = 144 - 720 = -576
Therefore, the required solutions are

Question. Solve the equation x2 - 2x + 3/2 = 0
The given quadratic equation is x2 - 2x + 3/2 = 0
This equation can also be written as 2x2 - 4x +3 = 0
On comparing this equation with ax2 + bx + c = 0 we obtain
a = 2, b = -4 and c = 3
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-4)2 - 4 × 2 × 3 = 16 - 24 = -8
Therefore, the required solutions are

Question. Solve the equation 27x2 - 10x + 1 = 0
The given quadratic equation is 27x2 - 10x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 27, b = -10, and c = 1
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-10)2 - 4 × 27 × 1 = 100 - 108 = -8
Therefore, the required solutions are

Question. Solve the equation 21x2 - 28x + 10 = 0
The given quadratic equation is 21x2 - 28x + 10 = 0
On comparing the given equation with ax2 + bx + c = 0 , we obtain
a = 21, b = -28 and c = 10
therefore, the discriminant of the given equation is
D = b2 - 4ac = (-28)2 - 4×21 ×10 = 784 - 840 = -56
Therefore, the required solutions are

Question. If z1 = 2 – i , z2 = 1+ i , find |(z1 + z2 + 1)/(z1 – z­2 + 1)|

z1 = 2 – I , z2 = 1+ i ,

Question. If (a + ib) = (x + 1)2 /(2x2 + 1) , prove that a2 + b2 = (x2 + 1)2 /(2x + 1)2

Question. Let z1 = 2 – i , z2 = -2+ i . find
(i) Re(z1 z2 /z1)
(ii) Im(1/z1 z1)

z1 = 2 – i , z2 = -2+ i
(i) z1 z2 = (2 - i)(-2 + i) = -4 + 2i +2i - i2  = -4 + 4i -(-1) = -3 + 4i

Question. Find the modulus and argument of the complex number (1 + 2i)/(1 - 3i).
Let z = (1 + 2i)/(1 - 3i) , then

Question. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Let z = (x - iy)(3 + 5i)
z = 3x + 5xi - 3yi - 5yi2 = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y)
∴ z  = (3x + 5y) - i(5x - 3y)
It is given that ,  z  = -6-24i
∴ (3x +5y) - i(5x - 3y) = -6 - 24i
Equating real and imaginary parts, we obtain
3x + 5y = -6 ...(i)
5x - 3y = 24 ...(ii)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

Putting the value of x in equation (i), we obtain
3(3) + 5y = -6
⇒ 5y = -6 - 9 = -15
⇒ y = -3
Thus, the values of x and y are 3 and -3  respectively.

Question. Find the modulus of [(1+i)/(1-i)] -[(1 -i)/(1 +i)].

Question. If α and β are different complex numbers with |β| = 1, then find |(β - α)/(1 - α β)|.
Let α = a + ib and β = x +iy
It is given that, |β| = 1
∴ √(x2 + y2 ) = 1
⇒ x2  + y2  = 1 ...(i)

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

Question. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that
(a2 + b2)(c2 + d2)(e2 + f2 )(g2 + h2) = A2 + B2

(a + ib)(c + id)(e + if)(g + ih) = A + iB
∴ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|
⇒ |(a + ib)|× |(c + id)|× |(e + if)| × |(g + ih)| = |A + B|  [|z1 z2| z1||z2|]

On squaring both sides, we obtain
(a2 + b2 )(c2 + d2 )(e2 + f2 )(g2 + h2 ) = A2 + B2 .
Hence, proved.

Question. If [(1 +i)/(1 - i)]m = 1, then find the least positive integral value of m.

∴ m = 4k, where k is some integer.
Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4× 1).

 NCERT Solutions Class 11 Mathematics Chapter 1 Sets
 NCERT Solutions Class 11 Mathematics Chapter 2 Relations and Functions
 NCERT Solutions Class 11 Mathematics Chapter 3 Trigonometric Functions
 NCERT Solutions Class 11 Mathematics Chapter 4 Principle of Mathematical Induction
 NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations
 NCERT Solutions Class 11 Mathematics Chapter 6 Linear Inequalities
 NCERT Solutions Class 11 Mathematics Chapter 7 Permutations and Combinations
 NCERT Solutions Class 11 Mathematics Chapter 8 Binomial Theorem
 NCERT Solutions Class 11 Mathematics Chapter 9 Sequences and Series
 NCERT Solutions Class 11 Mathematics Chapter 10 Straight Lines
 NCERT Solutions Class 11 Mathematics Chapter 11 Conic Sections
 NCERT Solutions Class 11 Mathematics Chapter 12 Introduction to Three Dimensional Geometry
 NCERT Solutions Class 11 Mathematics Chapter 13 Limits and Derivatives
 NCERT Solutions Class 11 Mathematics Chapter 14 Mathematical Reasoning
 NCERT Solutions Class 11 Mathematics Chapter 15 Statistics
 NCERT Solutions Class 11 Mathematics Chapter 16 Probability

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Yes, NCERT solutions for Class 11 Chapter 5 Complex Numbers and Quadratic Equations Mathematics are available in multiple languages, including English, Hindi