NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 11 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 11 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 11 Mathematics are an important part of exams for Class 11 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 11 Mathematics and also download more latest study material for all subjects. Chapter 5 Complex Numbers and Quadratic Equations is an important topic in Class 11, please refer to answers provided below to help you score better in exams

## 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: (5 i) (-3/5)i **

**Answer :**

(5

*i*) (-3/5)

*i*= -5 × (3/5) ×

*i × i*

= -3

*i*

^{2}

= -3(-1)

**[**

*i*

^{2}= -1]

= 3

**Question. Express the given complex number in the form a + ib : i^{9} + i^{19} **

**Answer :**

i

^{9}+ i

^{19}=

*i*

^{4×2+1}+

*i*

^{4×4+3}

= (

*i*

^{4})

^{2}.

*i*+ (

*i*

^{4})

^{4}.

*i*

^{3}

= 1×

*i*+ 1×(-

*i*) [

*i*

^{4}= 1,

*i*

^{3}= -

*i*]

= i + (-i)

= 0

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

**Answer :**

**Question. Express the given complex number in the form a + ib : 3(7 + i7) + i(7 + i7) **

**Answer : **

3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i^{2}

= 21 + 28i + 7 × (-1) **[∵i ^{2} = -1] **

= 14 + 28i

**Question. Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)**

**Answer :**

(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) **

**Answer :**

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

**Answer :**

**8. Express the given complex number in the form a + ib : (1 - i )^{4} **

**Answer :**

(1 -i)^{2} = [(1 - i)^{2}]^{2}

= [1^{2} + i^{2} + 2i^{2}]^{2}

= [1 - 1 - 2i]^{2}

= (-2i)^{2}

= (-2i)×(-2i)

= 4i^{2} = -4 **[i ^{2} = -1] **

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

**Answer :**

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

**Answer :**

**Question. Find the multiplicative inverse of the complex number 4 - 3i **

**Answer : **

Let z = 4 - 3i

Then, z = 4 + 3i and |z|^{2} = 4^{2} + (-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 **

**Answer : **

Let z = √5 + 3i

Then, z = √5 + 3i and |z|^{2} = (√5)^{2} + 3^{2} = 5 + 9= 14

Therefore, the multiplicative inverse of √5 + 3i is given by

**Question. Find the multiplicative inverse of the complex number – i**

**Answer :**

Let z = -i

Then, z = i and |z|

^{2}= 1

^{2}= 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)]**

**Answer :**

**Exercise 5.2**

**Question. Find the modulus and the argument of the complex number z = -1 - i√3 **

**Answer : **

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}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 3

⇒ r^{2} = 4 [cos^{2} θ + sin^{2} 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 **

**Answer : **

z = -√3 + i

Let r cos θ = -√3 and r sin θ = 1

On squaring and adding, we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = (-√3)^{2} + 1^{2}

⇒ r^{2} = 3 + 1 = 4 [cos^{2} θ + sin^{2} 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 **

**Answer : **

1 - i

Let r cos θ = 1 and r sin θ = -1

On squaring and adding, we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = 1^{2} + (-1)^{2}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 1

⇒ r^{2} = 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 **

**Answer : **

-1 + i

Let r cos θ = -1 and r sin θ = 1

On squaring and adding, we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = (-1)^{2} + 1^{2}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 1

⇒ r^{2} = 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 . **

**Answer : **

-1-i

Let r cos θ = -1 and r sin θ = -1

On squaring and adding, we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = (-1)^{2} + (-1)^{2}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 1

⇒ r^{2} = 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 **

**Answer : **

-3

Let r cos c = -3 and r sin θ = 0

On squaring and adding we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = (-3)^{2}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 9

⇒ r^{2} = 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 **

**Answer : **

√3 + i

let r cos θ = √3 and r sin θ = 1

On squaring and adding, we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = (√3)^{2} + 1^{2}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 3 + 1

⇒ r^{2} = 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**

**Answer : **

i

Let *r* cos*θ* = 0 and *r* sin *θ* = 1

On squaring and adding, we obtain

r^{2} cos^{2} θ + r^{2} sin^{2} θ = 0^{2} + 1^{2}

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 1

⇒ r^{2} = 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 x ^{2} + 3 = 0**

**Answer :**

The given quadratic equation is x

^{2}+ 3 = 0

On comparing the given equation with ax

^{2}+ bx + c = 0, we obtain

a = 1, b = 0, and c = 3

Therefore, the discriminant of the given equation is

D = b

^{2}– 4ac = 0

^{2}– 4 × 1 × 3 = –12

Therefore, the required solutions are

**Question. Solve the equation 2x ^{2} + x + 1 = 0**

**Answer :**

The given quadratic equation is 2

*x*

^{2}+

*x*+ 1 = 0

On comparing the given equation with

*ax*

^{2}+

*bx*+

*c*= 0, we obtain

*a*= 2,

*b*= 1, and

*c*= 1

Therefore, the discriminant of the given equation is

D =

*b*

^{2}– 4

*ac*= 1

^{2}– 4 × 2 × 1 = 1 – 8 = –7

Therefore, the required solutions are

**Question. ****Solve the equation x ^{2} + 3x + 9 = 0 **

**Answer :**

The given quadratic equation is

*x*

^{2}+ 3

*x*+ 9 = 0

On comparing the given equation with

*ax*

^{2}+

*bx*+

*c*= 0, we obtain

*a*= 1,

*b*= 3, and

*c*= 9

Therefore, the discriminant of the given equation is

D =

*b*

^{2}– 4

*ac*= 3

^{2}– 4 × 1 × 9 = 9 – 36 = –27

Therefore, the required solutions are

**Question. ****Solve the equation – x^{2} + x – 2 = 0**

**Answer :**

The given quadratic equation is –

*x*

^{2}+

*x*– 2 = 0

On comparing the given equation with

*ax*

^{2}+

*bx*+

*c*= 0, we obtain

*a*= –1,

*b*= 1, and

*c*= –2

Therefore, the discriminant of the given equation is

D =

*b*

^{2}– 4

*ac*= 1

^{2}– 4 × (–1) × (–2) = 1 – 8 = –7

Therefore, the required solutions are

**Question. ****Solve the equation x^{2} + 3x + 5 = 0**

**Answer :**

The given quadratic equation is

*x*

^{2}+ 3

*x*+ 5 = 0

On comparing the given equation with

*ax*

^{2}+

*bx*+

*c*= 0, we obtain

*a*= 1,

*b*= 3, and

*c*= 5

Therefore, the discriminant of the given equation is

D =

*b*

^{2}– 4

*ac*= 3

^{2}– 4 × 1 × 5 =9 – 20 = –11

Therefore, the required solutions are

**6. Solve the equation x^{2} – x + 2 = 0**

**Answer :**

The given quadratic equation is

*x*

^{2}–

*x*+ 2 = 0

On comparing the given equation with

*ax*

^{2}+

*bx*+

*c*= 0, we obtain

*a*= 1,

*b*= –1, and

*c*= 2

Therefore, the discriminant of the given equation is

D =

*b*

^{2}– 4

*ac*= (–1)

^{2}– 4 × 1 × 2 = 1 – 8 = –7

Therefore, the required solutions are

**Question. ****Solve the equation √2x ^{2} + x + √2 = 0 **

**Answer :**

The given quadratic equation is √2x

^{2}+ x + √2 = 0

On comparing the given equation with ax

^{2}+ bx + c = 0, we obtain

a = √2, b = 1, and c = √2

Therefore, the discriminant of the given equation is

D = b

^{2}- 4ac = 1

^{2}- 4 × √2 × √2 = 1 - 8 =-7

Therefore, the required solution are

**Question. ****Solve the equation √3x ^{2} - √2x + 3√3 = 0 **

**Answer :**

The given quadratic equation is √3x

^{2}- √2x + 3√3 = 0

On comparing the given equation with ax

^{2}+ bx + c = 0, we obtain

a = √3, b = -√2, and c = 3√3

Therefore, the discriminant of the given equation is

d = b

^{2}- 4ac = (-√2)

^{2}- 4(√3)(3√3) = 2 - 36 = -34

Therefore, the required solution are

**Question. ****Solve the equation x ^{2} + x + 1/√2 = 0 **

**Answer :**

The given quadratic equation is x

^{2}+ x+ 1/√2 = 0

This equation can also be written as √2x

^{2}+ √2x+ 1 = 0

On comparing this equation with ax

^{2}+ bx + c = 0, we obtain

a = √2, b = √2, and c = 1

∴ Discriminant (D) = b

^{2}- 4ac = (√2)

^{2}- 4×√2 ×1 = 2 - 4√2

Therefore, the required solutions are

**Question. ****Solve the equation x ^{2} + x/√2 + 1 = 0 **

**Answer :**

The given quadratic equation is x

^{2}+ x/√2 + 1 = 0

This equation can also be written as √2x

^{2}+ x + √2 = 0

On comparing this equation with ax

^{2}+bx + c = 0, we obtain

a = √2, b = 1, and c = √2

∴ Discriminant (D) = b

^{2}- 4ac = 1

^{2}- 4×√2×√2 = 1 - 8 = -7

Therefore, the required solutions are

**Miscellaneous Solutions**

**Question. Evaluate : [i ^{18} + (1/i)^{25} ]^{3} .**

**Answer :**

= [-1 - i]^{3}

= (-1)^{3} [1 +i]^{3}

= -[1^{3} + i^{3} + 3.1.i(1 + i)]

= -[1 + i^{3} + 3i + 3i^{2} ]

= -[1 - i + 3i -3]

= -[-2+2i]

= 2 -2i

**Question. For any two complex numbersz _{1} and z_{2}, prove that**

Re (z_{1}z_{2})= Re z_{1 }Re z_{2} – Im z_{1} Im z_{2}

**Answer :**

Let z

_{1}= x

_{1}+ iy

_{1}and z

_{2}= x

_{2}+ iy

_{2}

∴ z

_{1}z

_{2}= (x

_{1}+ iy

_{1})(x

_{2}+ iy

_{2})

= x

_{1}(x

_{2}+ iy

_{2}) + iy

_{1}(x

_{2}+ iy

_{2})

= x

_{1}x

_{2}+ ix

_{1}y

_{2}+ iy

_{1}x

_{2}+ i

^{2}y

_{1}y

_{2}

= x

_{1}x

_{2}+ ix

_{1}y

_{2}+ iy

_{1}x

_{2}+ y

_{1}y

_{2}

**[i**

^{2}= -1]= (x

_{1}x

_{2}- y

_{1}y

_{2}) + i(x

_{1}y

_{2}+ y

_{1}x

_{2})

⇒ Re (z

_{1}z

_{2}) = x

_{1}x

_{2}- y

_{1}x

_{2}

⇒ Re (z

_{1}z

_{2}) = Re z

_{1}Re z

_{2}- Im z

_{1}Im z

_{2}

Hence, proved.

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

**Answer : **

**Question. If x - iy = √(a - ib)/(c-id) prove that (x ^{2} + y^{2})^{2} = (a^{2} + b^{2})/(c^{2} + d^{2}) **

**Answer :**

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

**Answer :**

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain

r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 1

⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 2

⇒ r^{2} = 2 [cos^{2} θ + sin^{2} 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

r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 1

⇒r^{2} (cos^{2} θ + sin^{2} θ) = 2

⇒ r^{2} = 2 [cos^{2} θ + sin^{2} 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 3x ^{2} - 4x + 20/3 = 0 **

**Answer :**

The given quadratic equation is 3x

^{2}- 4x + 20/3 = 0

This equation can also be written as 9x

^{2}- 12x + 20 = 0

On comparing this equation with ax

^{2}+ bx + c = 0 , we obtain

a = 9 b = -12 and c = 20

Therefore, the discriminant of the given equation is

D = b

^{2}- 4ac = (-12)

^{2}- 4× 9 × 20 = 144 - 720 = -576

Therefore, the required solutions are

**Question. Solve the equation x ^{2} - 2x + 3/2 = 0 **

**Answer :**

The given quadratic equation is x

^{2}- 2x + 3/2 = 0

This equation can also be written as 2x

^{2}- 4x +3 = 0

On comparing this equation with ax

^{2}+ bx + c = 0 we obtain

a = 2, b = -4 and c = 3

Therefore, the discriminant of the given equation is

D = b

^{2}- 4ac = (-4)

^{2}- 4 × 2 × 3 = 16 - 24 = -8

Therefore, the required solutions are

**Question. ****Solve the equation 27x ^{2} - 10x + 1 = 0 **

**Answer :**

The given quadratic equation is 27x

^{2}- 10x + 1 = 0

On comparing the given equation with ax

^{2}+ bx + c = 0, we obtain

a = 27, b = -10, and c = 1

Therefore, the discriminant of the given equation is

D = b

^{2}- 4ac = (-10)

^{2}- 4 × 27 × 1 = 100 - 108 = -8

Therefore, the required solutions are

**Question. ****Solve the equation 21x ^{2} - 28x + 10 = 0 **

**Answer :**

The given quadratic equation is 21x

^{2}- 28x + 10 = 0

On comparing the given equation with ax

^{2}+ bx + c = 0 , we obtain

a = 21, b = -28 and c = 10

therefore, the discriminant of the given equation is

D = b

^{2}- 4ac = (-28)

^{2}- 4×21 ×10 = 784 - 840 = -56

Therefore, the required solutions are

**Question. ****If z _{1} = 2 – i , z_{2} = 1+ i , find |(z_{1} + z_{2} + 1)/(z_{1} – z_{2} + 1)|**

**Answer :**

z_{1} = 2 – I , z_{2} = 1+ i ,

**Question. ****If (a + ib) = (x + 1) ^{2} /(2x^{2} + 1) , prove that a^{2} + b^{2} = (x^{2} + 1)^{2} /(2x + 1)^{2} **

**Answer :**

**Question. ****Let z _{1} = 2 – i , z_{2} = -2+ i . find**

(i) Re(z_{1} z_{2} /z1)

(ii) Im(1/z_{1} z1)

**Answer :**

z

_{1}= 2 – i , z

_{2}= -2+ i

(i) z

_{1}z

_{2}= (2 - i)(-2 + i) = -4 + 2i +2i - i

^{2}= -4 + 4i -(-1) = -3 + 4i

**Question. ****Find the modulus and argument of the complex number (1 + 2i)/(1 - 3i).**

**Answer : **

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.**

**Answer :**

Let z = (x - iy)(3 + 5i)

z = 3x + 5xi - 3yi - 5yi

^{2}= 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)].
**

**Answer :**

**Question. If α and β are different complex numbers with |β| = 1, then find |(β - α)/(1 - α β)|. **

**Answer : **

Let α = a + ib and β = x +iy

It is given that, |β| = 1

∴ √(x^{2} + y^{2} ) = 1

⇒ x^{2} + y^{2} = 1 **...(i)**

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

**Answer :**

|1-x|

^{x}= 2

^{x}

**Question. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that
(a ^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2} )(g^{2} + h^{2}) = A^{2} + B^{2}. **

**Answer :**

(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|

**[|z**

_{1}z_{2}| z_{1}||z_{2}|]

On squaring both sides, we obtain

(a^{2} + b^{2} )(c^{2} + d^{2} )(e^{2} + f^{2} )(g^{2} + h^{2} ) = A^{2} + B^{2} .

Hence, proved.

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

**Answer :**

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

NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 11 Mathematics textbook online or you can easily download them in pdf.

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

The Class 11 Mathematics NCERT Solutions Chapter 5 Complex Numbers and Quadratic Equations are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 5 Complex Numbers and Quadratic Equations of Mathematics Class 11 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 5 Complex Numbers and Quadratic Equations Class 11 chapter of Mathematics so that it can be easier for students to understand all answers.

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

Class 11 Mathematics NCERT Solutions Chapter 5 Complex Numbers and Quadratic Equations is a really good source using which the students can get more marks in exams. The same questions will be coming in your Class 11 Mathematics exam. Learn the Chapter 5 Complex Numbers and Quadratic Equations questions and answers daily to get a higher score. Chapter 5 Complex Numbers and Quadratic Equations of your Mathematics textbook has a lot of questions at the end of chapter to test the students understanding of the concepts taught in the chapter. Students have to solve the questions and refer to the step-by-step solutions provided by Mathematics teachers on studiestoday to get better problem-solving skills.

**Chapter 5 Complex Numbers and Quadratic Equations Class 11 NCERT Solution Mathematics**

These solutions of Chapter 5 Complex Numbers and Quadratic Equations NCERT Questions given in your textbook for Class 11 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 11.

**Class 11 NCERT Solution Mathematics Chapter 5 Complex Numbers and Quadratic Equations**

NCERT Solutions Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations detailed answers are given with the objective of helping students compare their answers with the example. NCERT solutions for Class 11 Mathematics provide a strong foundation for every chapter. They ensure a smooth and easy knowledge of Revision notes for Class 11 Mathematics. As suggested by the HRD ministry, they will perform a major role in JEE. Students can easily download these solutions and use them to prepare for upcoming exams and also go through the Question Papers for Class 11 Mathematics to clarify all doubts

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