NCERT Solutions: Complex Numbers & Quadratic Equations

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
=

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:

 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 - i² = 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).