The chapter covers the basic principles and concepts of complex numbers and quadratic equations, and the solutions are structured in a concise and simple manner, making it easy for students to understand the concepts and score good marks in their exams. The NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations covers various topics such as complex numbers, algebra of complex numbers, quadratic equations, and many more.

**Exercise 5.1 : Solutions of Questions on Page Number : 103**

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

**Q1 :Express the given complex number in the form a + ib: **

**Answer :
**

**Q2 :Express the given complex number in the form a + ib: i ^{9} + i^{19
}**

**Answer :**

** **

**Q3 :Express the given complex number in the form a + ib: i ^{-39}**

**Answer :**

**Q4 :Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)**

**Answer :
**

**Q5 :Express the given complex number in the form a + ib: (1 – i) – (-1 + i6)
**

**Answer :**

**
**

**Q6 :Express the given complex number in the form a + ib: **

**Answer :
**

**Q7 :Express the given complex number in the form a + ib: **

**Answer :
**

**Q8 :Express the given complex number in the form a + ib: (1 – i) ^{4}**

**Answer**:

**Q9 :Express the given complex number in the form a + ib: **

**Answer :
**

**Q10 :Express the given complex number in the form a + ib: **

**Answer :
**

**Q11 :Find the multiplicative inverse of the complex number 4 – 3i**

**Answer :
**Let z = 4 – 3i

Then, = 4 + 3i and Therefore, the multiplicative inverse of 4 – 3i is given by

**Q12 :Find the multiplicative inverse of the complex number **

**Answer :
**Let z = Therefore, the multiplicative inverse of is given by

**Q13 :Find the multiplicative inverse of the complex number -i
**

**Answer :**

Let z = -i

Therefore, the multiplicative inverse of -i is given by

**Q14 :Express the following expression in the form of a + ib.**

** **

**Answer :
**

** Exercise 5.2 : Solutions of Questions on Page Number : 108**

**Q1 :Find the modulus and the argument of the complex number **

**Answer** :

On squaring and adding, we obtain

Since both the values of sin θand cos θ are negative and sinθand cosθare negative in III quadrant,

Thus, the modulus and argument of the complex number are 2 and respectively.

**Q2 :Find the modulus and the argument of the complex number **

**Answer :
**

On squaring and adding, we obtain

Thus, the modulus and argument of the complex number are 2 and respectively.

**Q3 :Convert the given complex number in polar form: 1 – i
**

**Answer :**

1 – i

Let rcos θ = 1 and rsin θ = -1

On squaring and adding, we obtain

This is the required polar form.

**Q4 :Convert the given complex number in polar form: – 1 + i
**

**Answer :**

-1 + i

Let rcos θ =-1 and rsin θ = 1

On squaring and adding, we obtain

It can be written,

This is the required polar form.

**Q5 :Convert the given complex number in polar form: – 1 – i
**

**Answer :**

– 1 – i

Let rcos θ = -1 and rsin θ = -1

On squaring and adding, we obtain

This is the required polar form.

**Q6 :Convert the given complex number in polar form: -3
**

**Answer :**

-3

Let rcos θ = -3 and rsin θ = 0

On squaring and adding, we obtain

This is the required polar form.

**Q7 :Convert the given complex number in polar form: **

**Answer :
**Let rcos θ =and rsin θ = 1

On squaring and adding, we obtain

This is the required polar form.

**Q8 : Convert the given complex number in polar form: i
Answer:
**i

Let rcosθ = 0 and rsin θ = 1

On squaring and adding, we obtain

This is the required polar form.

**Exercise 5.3 : Solutions of Questions on Page Number : 109**

** Q1 :Solve the equation x ^{2} + 3 = 0**

**Answer**:

The given quadratic equation is x

^{2}+ 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 = 0

^{2}– 4 × 1 × 3 = -12

Therefore, the required solutions are

**Q2 :Solve the equation 2x ^{2} + x + 1 = 0**

**Answer :**

The given quadratic equation is 2x

^{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}– 4ac = 12 – 4 × 2 × 1 = 1 – 8 = – 7

Therefore, the required solutions are

**Q3 :Solve the equation x ^{2} + 3x + 9 = 0**

**Answer :**

The given quadratic equation is x

^{2}+ 3x + 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}– 4ac = 32 – 4 × 1 × 9 = 9 – 36 = -27

Therefore, the required solutions are

**Q4 :Solve the equation -x ^{2} + x – 2 = 0**

**Answer :**

The given quadratic equation is -x2 + 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}– 4ac = 12 – 4 × (-1) × (-2) = 1 – 8 = -7

Therefore, the required solutions are

**Q5 :Solve the equation x ^{2} + 3x + 5 = 0**

**Answer :**

The given quadratic equation is x

^{2}+ 3x + 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}– 4ac = 32 – 4 × 1 × 5 =9 – 20 = -11

Therefore, the required solutions are

**Q6 :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}– 4ac = (-1)2 – 4 × 1 × 2 = 1 -“ 8 = -7

Therefore, the required solutions are

**Q7 :Solve the equation **

**Answer :
**The given quadratic equation is

On comparing the given equation with ax

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

a = , b = 1, and c =

Therefore, the discriminant of the given equation is

D = b^{2} – 4ac = 1^{2} – – = 1 – 8 = -7

Therefore, the required solutions are

**Q8 :Solve the equation **

**Answer :**

The given quadratic equation is

On comparing the given equation with ax^{2} + bx + c = 0, we obtain

a = , b =, and c =

Therefore, the discriminant of the given equation is

D = b^{2} – 4ac =

Therefore, the required solutions are

**Q9 :Solve the equation **

**Answer :**

The given quadratic equation is

This equation can also be written as

On comparing this equation with ax^{2} + bx + c = 0, we obtain

a = , b = , and c = 1

Therefore, the required solutions are

**Q10 :Solve the equation **

**Answer** :

The given quadratic equation is

This equation can also be written as

On comparing this equation with ax^{2} + bx + c = 0, we obtain

a = , b = 1, and c =

Therefore, the required solutions are

**Exercise Miscellaneous : Solutions of Questions on Page Number : 112**

** Q1 :Evaluate: **

**Answer :
**

**Q2 :For any two complex numbers z _{1} and z_{2}, prove that**

**Re (z**

_{1}z_{2}) = Re z_{1}Re z_{2}– Im z_{1}Im z_{2}**Answer :**

**Q3 :Reduce to the standard form. **

**Answer :
**

**Q4 :If x – iy = prove that **

**Answer :
**

**Q5 :Convert the following in the polar form:**

**(i) , (ii)
Answer :
(i)** Here,

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]

∴ z = r cos θ + i r sin θ

This is the required polar form

**(ii)** Here,

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]

∴z = r cos θ + i r sin θ

This is the required polar form.

**Q6 :Solve the equation **

**Answer :
**The given quadratic equation is

This equation can also be written as

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

**Q7 :Solve the equation **

**Answer :
**The given quadratic equation is

This equation can also be written as

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

**Q8 :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 = b2 – 4ac = (-10)2 – 4 × 27 × 1 = 100 – 108 = -8

Therefore, the required solutions are

**Q9 :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

**Q10 :If find **

**Answer :
**

Q11 :If find .

Answer :

**Q12 :If a + ib = , prove that a ^{2} + b^{2} = **

**Answer :**

On comparing real and imaginary parts, we obtain

Hence, proved.

**Q13 :Let . Find**

**(i) , (ii) **

**Answer :
**

(i)

On multiplying numerator and denominator by (2 – i), we obtain

On comparing real parts, we obtain

**(ii) **

On comparing imaginary parts, we obtain

**Q14 :Find the modulus and argument of the complex number .**

**Answer :**

Let , then

On squaring and adding, we obtain

Therefore, the modulus and argument of the given complex number are respectively.

**Q15 :Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of -6 – 24i.**

**Answer :
**Let

It is given that,

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

Thus, the values of x and y are 3 and -3 respectively.

**Q16 :Find the modulus of **

**Answer :
**

**Q17 :If (x + iy) ^{3} = u + iv, then show that **

**Answer :**

On equating real and imaginary parts, we obtain

Hence, proved.

=1

**Q19 :Find the number of non-zero integral solutions of the equation **

**Answer :
**Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.

**Q20 :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**:

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.

**Q21 :If , then find the least positive integral value of m.**

**Answer :
**

Therefore, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).