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

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Exercise 4.1

Que 1: Express the given complex number in the form a + ib : (5i)( -35 i)

Ans:
( -35 i) = -5 x -35 × i × i
= -3i²
= -3 × (-1)    [i² = -1]
= 3

Que 2: Express the given complex number in the form a + ib: i⁹+ i¹⁹

Ans:
i⁹ + i¹⁹ = i^4×2+1 + i^4×4+3
= (i⁴)² · i + (i⁴)⁴ · i³
= 1 × i + 1 × (−i) [i⁴ = 1, i³ = −i]
= i + (−i)
= 0

Que 3: Express the given complex number in the form a  + ib: i^–39

Ans:
i⁻³⁹ = i^−4×9−3 = (i⁴)⁻⁹ · i⁻³
= (1)⁻⁹ · i⁻³    [i⁴ = 1]
= ¹/_i³ = ¹/_−i    [i³ = −i]
= −¹/_i × i/i
= −ⁱ/_i² = −ⁱ/₋₁   [i² = −1]
= i

Que 4: Express the given complex number in the form a  + ib: 3(7 +  i7) +  i(7 +  i7)

Ans: 
3(7 + 7i) + i(7 + 7i) = 21 + 21i + 7i + 7i²
= 21 + 28i + 7 × (−1) [∵ i² = −1]
= 14 + 28i

Que 5: Express the given complex number in the form a +  ib: (1 – i) – (–1 + i6)

Ans:
(1 − i) − (1 + i6) = 1 − i + 1 − 6i
= 2 − 7i

Que 6: Express the given complex number in the form a  + ib: ( 15 + i 25) − ( 4 + i 52 )

Ans:
( 15 + i 25) − ( 4 + i 52 )
= 15 − 4 + i 25 − i 52
= ( 15 − 4) + i( 25 − 52)
= −195 + i −2110
= −195 − i 2110

Que 7: Express the given complex number in the form a +  ib:[ 13 + i 73 ] + [ 43 + i 13 ] − ( −43 + i)

Ans:
[ 13 + i 73 ] + [ 43 + i 13 ] − ( −43 + i)
= 13 + i 73 + 43 + i 13 + 43 − i
= ( 13 + 43 + 43) + i( 73 + 13 − 1)
= 173 + i 53

Que 8: Express the given complex number in the form a  + ib:  (1 – i)⁴

Ans:
(1 − i)⁴ = [(1 − i)²]²
= [1² + i² − 2i]²
= [1 − 1 − 2i]²
= (−2i)²
= (−2i) × (−2i)
= 4i² = −4        [∵ i² = −1]

Que 9: Express the given complex number in the form a  + ib: ( 13 + 3i)³

Ans:
( 13 + 3i)³ = 13³ + (3i)³ + 3 13(3i)( 13 + 3i)
= 127 + 27i³ + 3i( 13 + 3i)
= 127 + 27(−i) + i + 9i² [i³ = −i, i² = −1]
= 127 − 27i + i − 9
= ( 127 − 9) + i(−27 + 1)
= −24227 − 26i

Que 10: Express the given complex number in the form a +  ib: (−2 − 13 i)³

Ans:
(−2 − 13 i)³ = (−1)³(2 + 13 i)³
= −[ 2³ + (13 i)³ + 3(2)( 13 i )(2 + 13 i)]
= −[ 8 + i³27 + 2i( 2 + i3)]
= −[ 8 − i27 + 4i + 2i²3]      [i³ = −i, i² = −1]
= −[ 8 − i27 + 4i − 23]
= −[ 223 + i( 10727)]
= 223 − 10727 i

Que 11: Find the multiplicative inverse of the complex number: 4 – 3i

Ans:
Let z = 4 − 3i
Then, z̅ = 4 + 3i and |z|² = 4² + (−3)² = 16 + 9 = 25
Therefore, the multiplicative inverse of 4 − 3i is given by:
z⁻¹ = z̅|z|² = 4 + 3i25
= 425 + 325 i

Que 12: Find the multiplicative inverse of the complex number :√5 + 3i

Ans:
Let z = √5 + 3i
Then, z̅ = √5 − 3i and |z|² = (√5)² + 3² = 5 + 9 = 14
Therefore, the multiplicative inverse of √5 + 3i is given by:
z⁻¹ = z̅|z|² = √5 − 3i14
= √514 − 314i

Que 13: Find the multiplicative inverse of the complex number: –i

Ans: 
Let z = −i
Then, z̅ = i and |z|² = |i|² = 1² = 1
Therefore, the multiplicative inverse of −i is given by:
z⁻¹ = z̅|z|² = i1 = i

Que 14: Express the following expression in the form of a  + ib.

 (3 + i√5)(3 − i√5)(√3 + i√2) − (√3 − i√2)

Ans:
(3 + i√5)(3 − i√5)(√3 + i√2) − (√3 − i√2)
= (3)² − (i√5)²√3 + i√2 − √3 + i√2 [(a + b)(a − b) = a² − b²]
= 9 − 5i²2i√2
= 9 − 5(−1)2i√2 [i² = −1]
= 9 + 52i√2 = 14i2√2i
= 142(−1) = −7√2 − 7i2

Miscellaneous Exercise

Que 1: Evaluate: 

Ans:

 

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

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

Ans:
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
.∴ z₁z₂ = (x₁ + iy₁) (x₂ + iy₂)
= x₁(x₂ + iy₂) + iy₁(x₂ + iy₂)
= x₁x₂ + i(x₁y₂ + x₂y₁) + i²y₁y₂
= (x₁x₂ − y₁y₂) + i(x₁y₂ + x₂y₁)
.∴ Re(z₁z₂) = x₁x₂ − y₁y₂
.∴ Re(z₁z₂) = Re(z₁) Re(z₂) − Im(z₁) Im(z₂)
Hence, proved.

 Que 3: Reduce  to the standard form.

Ans:

 

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

 Que 4: If x – iy =  prove that .

Ans:

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

 (x² + y²)² = (x² − y²)² + 4x²y²
= (ac + bd)(c² + d²) + (ad − bc)(c² + d²)
= a²c² + b²d² + 2acbd + a²d² + b²c² − 2adbc(c² + d²)²
= a²c² + b²d² + a²d² + b²c² + 2acbd(c² + d²)²
= a²(c² + d²) + b²(c² + d²)(c² + d²)²
= (c² + d²)(a² + b²)
= (a² + b²)(c² + d²)
Hence, proved.

Que 5: If z₁ = 2 − i, z₂ = 1 + i, find z₁ + z₂ + 1z₁ − z₂ + i

Ans:

z₁ = 2 − i, z₂ = 1 + i
.∴ z₁ + z₂ + 1z₁ − z₂ + i = (2 − i) + (1 + i) + 1(2 − i) − (1 + i) + i

= 42(−i) = 2−i × 1 + i1 + i = 2(1 + i)1 − i²   [i² = −1]

= 2(1 + i)2 = 1 + i

|1 + i| = √(1² + 1²) = √2

Thus, the value of z₁ + z₂ + 1z₁ − z₂ + i is √2.

 Que 6: If a + ib = (x + i)²2x² + 1, prove that a² + b² = (x² + 1)²(2x² + 1)

Ans:

a + ib = (x + i)²2x² + 1

= x² + i² + 2xi2x² + 1

= x² − 1 + i2x2x² + 1

= x² − 12x² + 1 + i 2x2x² + 1

On comparing real and imaginary parts, we obtain
a = x² − 12x² + 1,   b = 2x2x² + 1

∴ a² + b² = (x² − 1)²(2x² + 1)² + (2x)²(2x² + 1)²

= x⁴ − 2x² + 1 + 4x²(2x² + 1)²

= x⁴ + 2x² + 1(2x² + 1)²

= (x² + 1)²(2x² + 1)²
∴ a² + b² = (x² + 1)²(2x² + 1)²
Hence, proved.

Que 7: Let z₁ = 2 − i, z₂ = −2 + i. Find:

(i) Re z₁z₂z₁
(ii) Im 1z₁z₁

Ans:

z₁ = 2 − i, z₂ = −2 + i
(i)
z₁z₂ = (2 − i)(−2 + i) = −4 + 2i + 2i − i² = −4 + 4i − (−1) = −3 + 4i

∴ z₁z₂z₁ = -3 + 4i2 + i

(ii)
On multiplying numerator and denominator by (2 − i), we obtain:

z₁z₂z₁ = (−3 + 4i)(2 − i)(2 + i)(2 − i)
= −6 + 3i + 8i − 4i²2² + 1²
= −6 + 11i − 4(-1)2² + 1²
= −2 + 11i5
= = −25 + 11i5
 On comparing real parts, we obtain

 

(ii)

On comparing imaginary parts, we obtain

 

Que 8: Find the real numbers x and y if (x – iy) (3+5i) is the conjugate of –6 – 24i.

Ans:
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,

∴ (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:

9x + 15y = −18   ... (from i × 3)
25x − 15y = 120   ... (from ii × 5)

Adding: 34x = 102

∴ x = 10234 = 3

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.

Que 9: Find the modulus of 1 + i1 − i − 1 − i1 + i.

Ans:
1 + i1 − i − 1 − i1 + i = (1 + i)² − (1 − i)²(1 − i)(1 + i)
= 1 + i² + 2i − 1 − i² − 2i1² + 1² = 4i2 = 2i
∴ 1 + i1 − i − 1 − i1 + i = |2i| = √2² = 2

Que 10: If (x + iy)³ = u + iv, then show that ux + vy = 4(x² − y²).

Ans:

(x + iy)³ = u + iv

⇒ x³ + (iy)³ + 3·x·iy(x + iy) = u + iv

⇒ x³ + i y³ + 3x²yi + 3xy²i² = u + iv

⇒ x³ − 3xy² + i(3x²y − y³) = u + iv

∴ u = x³ − 3xy², v = 3x²y − y³

On equating real and imaginary parts, we obtain:

∴ ux + vy = x³ − 3xy²x + 3x²y − y³y

= x(x² − 3y²)x + y(3x² − y³)y

= x² − 3y² + 3x² − y²

= 4x² − 4y²

= 4(x² − y²)

∴ ux + vy = 4(x² − y²)

Hence, proved.

Que 11: If α and β are different complex numbers with  = 1, then find .

Ans:

Let α = a  + ib and β = x +  iy

It is given that,

 

 

 

Que 12: Find the number of non-zero integral solutions of the equation .

Ans:
|1 − i|^x = 2^x

⇒ √(1² + (−1)²)^x = 2^x

⇒ (√2)^x = 2^x

⇒ 2^x/2 = 2^x

⇒ x 2 = x

⇒ x = 2x

⇒ 2x − x = 0

⇒ x = 0

 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.

Que 13: If (a  + ib) (c +  id) (e +  if) (g  + ih) = A + iB, then show that

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

 Ans:

(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 + iB|

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

On squaring both sides, we obtain:

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

Hence, proved.

Que 14: If (1 + i)(1 − i)^m = 1, then find the least positive integral value of m.

Ans:

(1 + i)(1 − i)^m = 1

⇒ (1 + i)(1 − i) × (1 + i)(1 + i)^m = 1

⇒ (1 + i)²(1² + 1²)^m = 1

⇒ (1² + i² + 2i)2^m = 1

⇒ (−1 + 2i)2^m = 1

⇒ (2i)2^m = 1

⇒ i^m = 1

∴ 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).

 Que 15: Convert the following in the polar form (Out of NCERT)

(i) 1 + 7i(2 − i)²,

(ii) 1 + 3i(1 − 2i) 

Ans:

 (i) Here, z = 1 + 7i(2 − i)²

 = 1 + 7i(2 − i)² = 1 + 7i4 + i² − 4i = 1 + 7i4 − 1 − 4i
= 1 + 7i3 − 4i × 3 + 4i3 + 4i = 3 + 4i + 21i + 28i²3² + 4²
= 3 + 4i + 21i − 283² + 4² = −25 + 25i25
= −1 + i

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 cos θ = −1 and √2 sin θ = 1

⇒ cos θ = −1√2, sin θ = 1√2

.∴ θ = π − π4 = 3π4   [As θ lies in II quadrant]

.∴ z = r cos θ + i r sin θ

= √2 cos 3π4 + i √2 sin 3π4

= √2 ( cos 3π4 + i sin 3π4 )

 This is the required polar form.

(ii) Here, z =  1 + 3i(1 − 2i)

 = 1 + 3i1 − 2i × 1 + 2i1 + 2i

= 1 + 2i + 3i − 61 + 4

= −5 + 5i5 = −1 + i

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 cos θ = −1 and √2 sin θ = 1

⇒ cos θ = −1√2, sin θ = 1√2

.∴ θ = π − π4 = 3π4   [As θ lies in II quadrant]

.∴ z = r cos θ + i r sin θ

= √2 cos 3π4 + i √2 sin 3π4

= √2 ( cos 3π4 + i sin 3π4 )

This is the required polar form.

Que 16: Solve the equation: 3x² − 4x + 203 = 0  (Out of NCERT)

Ans:

The given quadratic equation is 3x² − 4x + 203 = 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:
x = −b ± √D2a
= −(−12) ± √−576
2 × 9
= 12 ± √576 i18   [√−1 = i]

= 12 ± 24i18
= 6(2 ± 4i)18
= 2 ± 4i3
= 23 ± 4i3

 Que 17: Solve the equation : x² − 2x + 32 = 0 (Out of NCERT)

Ans:

The given quadratic equation is x² − 2x + 32 = 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:
x = −b ± √D2a
= −(−4) ± √−82 × 2
= 4 ± 2√2i4   [√−1 = i]

= 2 ± √2i2
= 1 ± √2i2

Que 18: Solve the equation: 27x² – 10x + 1 = 0 (Out of NCERT)

Ans:

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

 x = −b ± √D2a = −(−10) ± √−82 × 27 = 10 ± 2√2i54   [√−1 = i]

= 5 ± √2i27 = 527 ± √2i27

 Que 19: Solve the equation: 21x² – 28x + 10 = 0 (Out of NCERT)

Ans:

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

 = −b ± √D2a = −(−28) ± √−562 × 21 = 28 ± √56 i42

= 28 ± 2√14 i42 = 2842 ± 2√1442 = 23 ± √1421 i

Que 20: Find the modulus and argument of the complex number 1 + 2i1 − 3i. (Out of NCERT)

Ans:

Let z = 1 + 2i1 − 3i, then:

z = (1 + 2i)(1 + 3i)(1 − 3i)(1 + 3i) = 1 + 3i + 2i + 6i²1² + 3² = 1 + 5i + 6(−1)1 + 9

= −5 + 5i10 = −510 + 5i10 = −12 + 12i

Let z = r cosθ + i r sinθ
i.e., r cosθ = −12 and r sinθ = 12

On squaring and adding, we obtain:

r²(cos²θ + sin²θ) = (−1)²2² + (1)²2² = 14 + 14 = 12

⇒ r² = 12, ⇒ r = 1√2 [Conventionally, r > 0]

∴ 1√2 cosθ = −12, and 1√2 sinθ = 12

⇒ cosθ = −1√2, sinθ = 1√2

⇒ θ = π − π4 = 3π4 [As θ lies in the II quadrant]

 Therefore, the modulus and argument of the given complex number are  1√2 and 3π4  respectively.