NCERT Solutions Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

Exercise 4.1:

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

Q2: Express the given complex number in the form a + ib: i⁹+ i¹⁹
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Q3: Express the given complex number in the form a  + ib: i^–39
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Q4: Express the given complex number in the form a  + ib: 3(7 +  i7) +  i(7 +  i7)
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Q5: Express the given complex number in the form a +  ib: (1 – i) – (–1  i6)
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Q6: Express the given complex number in the form a  + ib:
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Q7: Express the given complex number in the form a +  ib:
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Q8: Express the given complex number in the form a  + ib:  (1 – i)⁴
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Q9: Express the given complex number in the form a  + ib:
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Q10: Express the given complex number in the form a +  ib:
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Q11: Find the multiplicative inverse of the complex number 4 – 3i
Ans:
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 
Ans:
Let z =

 
Therefore, the multiplicative inverse of  is given by

 

Q13: Find the multiplicative inverse of the complex number –i
Ans: Let z = –i

 
Therefore, the multiplicative inverse of –i is given by

 

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

 
Ans:

Exercise Miscellaneous

Question 1: Evaluate: 

Answer:

Question 2: For any two complex numbers z₁ and z₂, prove that

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Answer:

Question 3: Reduce  to the standard form.

Answer:

 [ On Multiplying numerator and denomunator by (14 + 5i)

Question 4: If x – iy =  prove that .

Answer:

 [ On Multiplying numerator and denomunator by (c + id)]

Question 5: 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

r² (cos² θ + sin² θ) = 1+ 1

⇒ r² (cos² θ + sin² θ) = 2
⇒ r² = 2     [cos² θ + sin² 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

r² (cos² θ + sin² θ) = 1+ 1
⇒r² (cos² θ  +sin² θ) = 2

⇒ r² = 2                        [cos² θ + sin² θ = 1]

∴z = r cos θ  + i r sin θ

This is the required polar form.

Question 6: Solve the equation

Answer:

The given quadratic equation is

This equation can also be written as

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 7: Solve the equation

Answer:

The given quadratic equation is

This equation can also be written as

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 8: Solve the equation 27x² – 10x + 1 = 0

Answer:

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 9: Solve the equation 21x² – 28x + 10 = 0

Answer:

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 9: If   find .

Answer:

Question 10: If a +  ib = , prove that a² +  b² =

Answer:

On comparing real and imaginary parts, we obtain

Hence, proved.

Question 10: 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

Question 11: 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.

Question 12: 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

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.

Question 13: Find the modulus of  .

Answer:

Question 14: If (x  + iy)³ = u  + iv, then show that .

Answer:

On equating real and imaginary parts, we obtain

Hence, proved.

Question 15: If α and β are different complex numbers with  = 1, then find .

Answer:

Let α = a  + ib and β = x +  iy

It is given that,

Question 16: 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.

Question 17: If (a  + ib) (c +  id) (e +  if) (g  + ih) = A + iB, then show that

(a²  + b²) (c²+   d²) (e² +  f²) (g² +  h²) = A²  +B².

Answer:

On squaring both sides, we obtain

(a²  + b²) (c² +  d²) (e² +  f²) (g²  + h²) = A² + B²

Hence, proved.

Question 18: 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).