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

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 5
Chapter Name : Complex Number And Quadratic Equations

Exercise 5.1

Q1 Express each of the complex number in the form a + ib.
3
(5i) (− i)
5

−3 3
(5i) ( i) = −5 × × i × i
5 5

2
= −3i

= −3(−1)

= 3

Page : 103 , Block Name : Exercise 5.1

Q2 Express each of the complex number in the form a + ib.
9 19
i + i

9 19 4x−1 4x+1
i + i = i + i
2 4
4 4 3
= (i ) ⋅ j + (j ) ⋅ i

= 1 × i + | × (−i)

= i + (−i)

= 0

Page : 104 , Block Name : Exercise 5.1

Q3 Express each of the complex number in the form a + ib.
−39
i

−9
−14 −1×4−3 4 −3
I = f = (i ) ⋅ l

−9 −3
= (1) − F

1 1
= =
3 −i
p

−1 j
= ×
i i

−i −i
= = = i
2 −1
i

Page : 104 , Block Name : Exercise 5.1

Q4 Express each of the complex number in the form a + ib.
3(7 + i7) + i(7 + i7)

2
3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i

= 21 + 28i + 7 × (−1)

= 14 + 28i

Page : 104 , Block Name : Exercise 5.1

Q5 Express each of the complex number in the form a + ib.
(1 − i) − (−1 + i6)

(1 − i) − (−1 + i6) = 1 − j + 1 − 6i

= 2 − 7i

Page : 104 , Block Name : Exercise 5.1

Q6 Express each of the complex number in the form a + ib.
1 2 5
( + i ) − (4 + i )
5 5 2

1 2 5
( + i ) − (4 + i )
5 5 2

1 2 5
= + i − 4 − l
5 5 2

1 2 5
= ( − 4) + i ( − )
5 5 2

−19 −21
= + i( )
5 10

−19 21
= − i
5 10

Page : 104 , Block Name : Exercise 5.1

Q7 Express each of the complex number in the form a + ib.
1 7 1 4
[( + i ) + (4 + i )] − (− + i)
3 3 3 3

1 7 1 −4
[( + i ) + (4 + i )] − ( + i)
3 3 3 3

1 7 1 4
= + i + 4 + i + − i
3 3 3 3

1 4 7 1
= ( + 4 + ) + ( + − 1)
3 3 3 3

17 5
= + i
3 3

Page : 104 , Block Name : Exercise 5.1

Q8 Express each of the complex number in the form a + ib.
4
(1 − i)

2
t 2
(1 − i) = [(1 − i) ]

2
2 2
= [1 + i − 2i]

2
= [1 − 1 − 2i]

2
= (−2i)

= (−2i) × (−2i)

2
= 4i = −4

Page : 104 , Block Name : Exercise 5.1

Q9 Express each of the complex number in the form a + ib.
3
1
( + 3i)
3

3 3
1 1 1 1
3
( + 3i) = ( ) + (3i) + 3( ) (3i) ( + 3i)
3 3 3 3

1 1
3
= + 27i + 3i ( + 3i)
27 3

1
2
= + 27(−i) + i + 9i
27

1
= − 27i + i − 9
27

1
= ( − 9) + i(−27 + 1)
27

−242
= − 26i
27

Page : 104 , Block Name : Exercise 5.1

Q10 Express each of the complex number in the form a + ib.
3
1
(−2 − i)
3

3 3
1 1
3
(−2 − i) = (−1) (2 + i)
3 3

2
i i i
3
= − [2 + ( ) + 3(2) ( ) (2 + )]
3 3 3

3
i i
= − [8 + + 2i (2 + )]
27 3
2
i 2i
= − [8 − + 4i + ]
27 3

i 2
= − [8 − + 4i − ]
27 3

22 107i
= −[ + ]
3 27

22 107
= − − i
3 27

Page : 104 , Block Name : Exercise 5.1

Q11 Find the multiplicative inverse of each of the complex numbers given
4 − 3i

Let z = 4 − 3i

2 2 2
¯
¯¯
Then, z = 4 + 3i and |z| = 4 + (−3) = 16 + 9 = 25

Therefore, the multiplicative inverse of 4 − 3i is given by

−1 z 4+3i 4 3
z = 2
= = + i
25 25 25
|z|

Page : 104 , Block Name : Exercise 5.1

Q12 Find the multiplicative inverse of each of the complex numbers given
√5 + 3i

Let z = √5 + 3i

2 2 2
¯
¯¯
Then, z = √5 − 3i and |z| = (√5) + 3 = 5 + 9 = 14

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

¯
¯
z
¯ √5−3i √5 3i
−1
z = = = −
2
|z| 14 14 14

Page : 104 , Block Name : Exercise 5.1

Q13 Find the multiplicative inverse of each of the complex numbers given
−i

Let z = −i

2 2
¯
¯¯
Then, z = i and |z| = 1 = 1

Therefore, the multiplicative inverse of − i is given by

¯
¯¯
−1 z i
z = 2
= = i
1
|z|

Page : 104 , Block Name : Exercise 5.1

Q14 Express the following expression in the form of a + ib
(3+i√5)(3−i√5)

(√3+√2i)−(√3−i√2)

(3+i√5)(3−i√5)

(√3+√2i)−(√3−i√2)

2 2
(3) −(i√5)
=
√3+√2i−√3+√2i

2
9−5i
=
2√2i

9−5(−1)
=
2√2i

9+5 i
= ×
i
2√2i

14i
=
2
2√2i

14i
=
2√2(−1)

−7i √2
= ×
√2 √2

−7√2i
=
2

Page : 104 , Block Name : Exercise 5.1

Exercise 5.2

Q1 Find the modulus and the arguments of each of the complex numbers
z = −1 − i√3

z = −1 − i√3

Let r cos θ = −1 and r sin θ = −√3

On squaring and adding, we obtain

2 2 2 2
(r cos θ) + (r sin θ) = (−1) + (−√3)

2 2 2
⇒ r (cos θ + sin θ) = 1 + 3

⇒ Modulus = 2

∴ 2 cos θ = −1 and 2 sin θ = −√3

−1 −√3
⇒ cos θ = and sin θ =
2 2

since both the values of sin θ and cos θ are negative and sin θ and cos θ are negative in

III quadrant,

π −2π
Argument = − (π − ) =
3 3

−2π
Thus, the modulus and argument of the complex number − 1 − √3i are 2 and
3

respectively.

Page : 108 , Block Name : Exercise 5.2

Q2 Find the modulus and the arguments of each of the complex numbers z = −√3 + i

z = − √3 + i

Let r cos θ = −√3 and r sin θ = 1

On squaring and adding, we obtain

2 2 2 2 2 2
r cos θ + r sin θ = (−√3) + 1

2
⇒ r = 3 + 1 = 4

⇒ r = √4 = 2

∴ Modulus = 2

∴ 2 cos θ = −√3 and 2 sin θ = 1

−√3 1
⇒ cos θ = and sin θ =
2 2

π 5π
∴ θ = π − =
6 6

5π
Thus, the modulus and argument of the complex number − √3 + i are 2 and
6

respectively.

Page : 108 , Block Name : Exercise 5.2

Q3 Convert each of the complex numbers given in the polar form:
1 − i

1 − i

Let r cos θ = 1 and r sin θ = −1

On squaring and adding, we obtain
2 2 2 2 2 2
r cos θ + r sin θ = 1 + (−1)

2 2 2
⇒ r (cos θ + sin θ) = 1 + 1

2
⇒ r = 2

⇒ r = √2

∴ √2 cos θ = 1 and √2 sin θ = −1

1 1
⇒ cos θ = and sin θ = −
√2 √2

π π π π
∴ 1 − i = r cos θ + r sin θ = √2 cos(− ) + i√2 sin(− ) = √2 [cos(− ) + i sin(− )] This is
4 4 4 4

the required polar form.

Page : 108 , Block Name : Exercise 5.2

Q4 Convert each of the complex numbers given in the polar form:
−1 + i

−1 + i

Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

2 2 2 2 2 2
r cos θ + r sin θ = (−1) + 1

2 2 2
⇒ r (cos θ + sin θ) = 1 + 1

⇒ r = √2 [ Conventionally, r > 0]

∴ √2 cos θ = −1 and √2 sin θ = 1

1 1
⇒ cos θ = − and sin θ =
√2 √2

π 3π
∴ θ = π − = [ As θ lies in the II quadrant ]
4 4

It can be written,

3π 3π 3π 3π
∴ −1 + i = r cos θ + ir sin θ = √2 cos + i√2 sin = √2 (cos + i sin )
4 4 4 4

This is the required polar form.

Page : 108 , Block Name : Exercise 5.2

Q5 Convert each of the complex numbers given in the polar form:
−1 − i

−1 − i

Let r cos θ = −1 and r sin θ = −1

On squaring and adding, we obtain

2 2 2 2 2 2
r cos θ + r sin θ = (−1) + (−1)

2 2 2
⇒ r (cos θ + sin θ) = 1 + 1

2
⇒ r = 2

⇒ r = √2

∴ √2 cos θ = −1 and √2 sin θ = −1

1 1
⇒ cos θ = − and sin θ = −
√2 √2

π 3π
∴ θ = − (π − ) = − [ As θ lies in the III quadrant ]
4 4

−3π −3π −3π −3π
∴ −1 − i = r cos θ + ir sin θ = √2 cos + i√2 sin = √2 (cos + i sin ) This is the
4 4 4 4

required polar form.

Page : 108 , Block Name : Exercise 5.2

Q6 Convert each of the complex numbers given in the polar form:
−3

Let r cos θ = −3 and r sin θ = 0

On squaring and adding, we obtain
2 2 2 2 2
r cos θ + r sin θ = (−3)

2 2 2
⇒ r (cos θ + sin θ) = 9

2
⇒ r = 9

⇒ r = √9 = 3

∴ 3 cos θ = −1 and 3 sin θ = 0

⇒ θ = π

∴ −3 = r cos θ + ir sin θ = 3 cos π + B sin π = 3(cos π + i sin π)

This is the required polar form.

Page : 108 , Block Name : Exercise 5.2

Q7 Convert each of the complex numbers given in the polar form:
√3 + i

√3 + i

Let r cos θ = √3 and r sin θ = 1

On squaring and adding, we obtain

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

2 2 2
⇒ r (cos θ + sin θ) = 3 + 1

2
⇒ r = 4

⇒ r = √4 = 2 [Conventionally, r > 0]

∴ 2 cos θ = √3 and 2 sin θ = 1

√3 1
⇒ cos θ = and sin θ =
2 2

π
∴ θ =
6

π π π π
∴ √3 + i = r cos θ + ir sin θ = 2 cos + i2 sin = 2 (cos + i sin )
6 6 6 6

Page : 108 , Block Name : Exercise 5.2

Q8 Convert each of the complex numbers given in the polar form:
i

Let r cos θ = 0 and r sin θ = 1

On squaring and adding, we obtain

2 2 2 2 z 2
r cos θ + r sin θ = 0 + 1

2 2 2
⇒ r (cos θ + sin θ) = 1

2
⇒ r = 1

⇒ r = √1 = 1

∴ cos θ = 0 and sin θ = 1
π
∴ θ =
2

π π
∴ i = r cos θ + ir sin θ = cos + i sin
2 2

This is the required polar form.

Page : 108 , Block Name : Exercise 5.2

Exercise 5.3

Q1 Solve each of the following equations:
2
x + 3 = 0

2
The given quadratic equation is x + 3 = 0

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

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

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = 0 − 4 × 1 × 3 = −12

Therefore, the required solutions are

−b ± √D ±√−12 ±√12i
= =
2a 2 × 1 2

±2√3i
= = ±√3i
2

Page : 109 , Block Name : Exercise 5.3

Q2 Solve each of the following equations:
2
2x + x + 1 = 0

2
The given quadratic equation is 2x + x + 1 = 0

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

a = 2, b = 1, and c = 1

Ther efore, the discriminant of the given equation is

2 2
D = b − 4ac = 1 − 4 × 2 × 1 = 1 − 8 = −7

Therefore, the requlred solutlons are

−b±√D −1±√−7 −1±√7i
= =
2a 2×2 4

Page : 109 , Block Name : Exercise 5.3

Q3 Solve each of the following equations:
2
x + 3x + 9 = 0

2
The given quadratic equation is x + 3x + 9 = 0

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

a = 1, b = 3, and c = 9

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = 3 − 4 × 1 × 9 = 9 − 36 = −27

Ther efore, the required solutions are

−b±√D −3±√−27 −3±3√−3 −3±3√3i
= = =
2a 2(1) 2 2

Page : 109 , Block Name : Exercise 5.3

Q4 Solve each of the following equations:
2
−x + x − 2 = 0

2
The given quadratic equation is − x + x − 2 = 0

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

a = −1, b = 1, and c = −2

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = 1 − 4 × (−1) × (−2) = 1 − 8 = −7

Therefore, the required solutions are

−b±√D −1±√−7 −1±√7i
= =
2a 2×(−1) −2

Page : 109 , Block Name : Exercise 5.3

Q5 Solve each of the following equations:
2
x + 3x + 5 = 0

2
The given quadratic equation is x + 3x + 5 = 0

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

a = 1, b = 3, and c = 5

Therefore, the discriminant of the glven equation is

2 2
D = b − 4ac = 3 − 4 × 1 × 5 = 9 − 20 = −11

Therefore, the requlred solutions are

−b±√D −3±√−11 −3±√11i
= =
2a 2×1 2

Page : 109 , Block Name : Exercise 5.3

Q6 Solve each of the following equations:
2
x − x + 2 = 0

2
The given quadratic equation is x − x + 2 = 0

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

a = 1, b = −1, and c = 2

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = (−1) − 4 × 1 × 2 = 1 − 8 = −7

Therefore, the required solutions are

−b±√D −(−1)±√−7 1±√7i
= =
2a 2×1 2

Page : 109 , Block Name : Exercise 5.3

Q7 Solve each of the following equations:
2
√2x + x + √2 = 0

2
The given quadratic equation is √2x + x + √2 = 0

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

a = √2, b = 1, and c = √2

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = 1 − 4 × √2 × √2 = 1 − 8 = −7

Therefore, the required solutions are

−b±√D −1±√−7 −1±√7i
= =
2a
2×√2 2√2

Page : 109 , Block Name : Exercise 5.3

Q8 Solve each of the following equations:
2
√3x − √2x + 3√3 = 0

2
The given quadratic equation is √3x − √2x + 3√3 = 0

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

a = √3, b = −√2, and c = 3√3

Therefore, the discriminant of the given equation is
2 2
D = b − 4ac = (−√2) − 4(√3)(3√3) = 2 − 36 = −34

Therefore, the required solutions are

−b±√D −(−√2)±√−34 √2±√34i
= = [√−1 = i]
2a
2×√3 2√3

Page : 109 , Block Name : Exercise 5.3

Q9 Solve each of the following equations:
2 1
x + x + = 0
√2

The given quadratic equation is

2
This equation can also be written as √2x + √2x + 1 = 0

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

a = √2, b = √2, and c = 1

2 2
∴ Discriminant (D) = b − 4ac = (√2) − 4 × (√2) × 1 = 2 − 4√2

Therefore, the required solutions are

−√2 ± √2(1 − 2√2)
−b ± √D −√2 ± √2 − 4√2
= =
2a −1±(√2√2−1)i
2 × √2 2 √2
=
2
−√2 ± √2(√2√2 − 1)
= ( )
2 √2

Page : 109 , Block Name : Exercise 5.3

Q10 Solve each of the following equations:
2 x
x + + 1 = 0
√2

The given quadratic equation is

2
This equation can also be written as √2x + x + √2 = 0

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

a = √2, b = 1, and c = √2

2 2
∴ Discriminant (D) = b − 4ac = 1 − 4 × √2 × √2 = 1 − 8 = −7

Therefore, the required solutions are

−b±√D −1±√−7 −1±√7i
= = [√−1 = i]
2a
2√2 2√2

Page : 109 , Block Name : Exercise 5.3

Miscellaneous Exercise

Q1 Evaluate
3
25
18 1
[i + ( ) ]
i

3
25
18 1
[i + ( ) ]
i

3
1
4x+42
= [i + ]
4x+1
i
3

4 1
4 2
= [(i ) − i + ]
6
4
(i ) ⋅ i
3
1
2
= [t + ]
i

3
1 i
= [−1 + × ]
i i
3
j
= [−1 + ]
i2

3
j
= [−1 + ]
2
j
3
= [−1 − i]

3 3
= (−1) [1 + i]
5 2
= − [1 + i + 3i + 3i ]

= [1 − i + 3i − 3]

= −[−2 + 2i]

= 2 − 2i

Page : 112 , Block Name : Miscellaneous Exercise

Q2
For any two complex numbers z1 and z2 , prove that

Re(z1 z2 ) = Re z1 Re z2 − Im z1 Im z2

Let z1 = x1 + iy1 and z2 = x2 + iy2

∴ z1 z2 = (x1 + iy1 ) (x2 + iy2 )

= x1 (x2 + iy2 ) + iy1 (x2 + ty2 )

2
= x1 x2 + ix1 y2 + iy1 x2 + i y1 y2

= x1 x2 + ix1 y2 + iy1 x2 − y1 y2

= (x1 x2 − y1 y2 ) + i (x2 y2 + y1 x2 )

⇒ Re(z1 z2 ) = x1 x2 − y1 y2

⇒ Re(z1 z2 ) = Re z1 Re z2 − lm z1 Im z2

Hence, proved.

Page : 112 , Block Name : Miscellaneous Exercise

Q3 Reduce ( to standard form
1 2 3−4i
− )( )
1−4i 1+i 5+i

1 2 3−4i (1+i)−2(1−4i) 3−4i
( − )( ) = [ ][ ]
1−4i 1+i 5+i (1−4i)(1+i) 5+i

1+i−2+8i 3−4i −1+9i 3−4i
= [ ][ ] = [ ][ ]
1+i−4i−4i2 5+i 5−3i 5+i

2
−3+4i+27i−36i 33+31i 33+31i
= [ ] = =
25+5i−15i−3i2 28−10i 2(14−5i)

(33+31i) (14+5i)
= × [ On multiplying numerator and denominator by (14 + 5i)]
2(14−5i) (14+5i)

2
462 + 165i + 434i + 155i 307 + 599i
= =
2 2 2
2 [(14) − (5i) ] 2 (196 − 25i )

307 + 599i 307 + 599i 307 599i
= = = +
2(221) 442 442 442

Page : 112 , Block Name : Miscellaneous Exercise

2 2 2

Q4 if x − iy = √
a−ib 2 2 a +b
prove that (x + y ) = 2 2
c−id c +d

a − ib
x − iy = √
c − id

a − ib c + id
= √ × [On multiplying numerator and denomin ator by (c + id)]
c − id c + id

(ac+bd)+i(ad−bc)
= √ 2 2
c +d

(ac+bd)+i(ad−bc)
2
∴ (x − iy) = 2 2
c +d

(ac+bd)+i(ad−bc)
2 2
⇒ x − y − 2ixy = 2 2
c +d

(ac+bd)+i(ad−bc)
2
∴ (x − iy) = 2 2
c +d

(ac+bd)+i(ad−bc)
2 2
⇒ x − y − 2ixy = 2 2
c +d

2 2 ac+bd ad−bc
x − y = 2 2
, −2xy = 2 2
c +d c +d

2 2
2 2 2 2 2 2
(x + y ) = (x − y ) + 4x y

2 2 [Using (i)]
ac+bd ad−bc
= ( ) + ( )
c2 +d 2 c2 +d 2

2 2 2 2 2 2 2 2
a c + b d + 2acbd + a d + b c − 2adbc
=
2
2 2
(c + d )

2 2 2 2 2 2 2 2
a c + b d + a d + b c
=
2
2 2
(c + d )

2 2 2 2 2 2
a (c + d ) + b (c + d )
=
2
2 2
(c + d )
2 2 2 2
(c +d )(a +b )

= 2
(c2 +d 2 )

2 2
a +b
= 2 2
c +d

Hence, proved.

Page : 112 , Block Name : Miscellaneous Exercise

Q5
Convert the following in the polar form:

1+7i 1+3i
(i) 2
, (ii)
(2−i) 1−2i

1+7i
(i) Here z = 2
(2−i)

1+7i 1+7i 1+7i
= 2
= 2
=
(2−i) 4+i −4i 4−1−4i

2
1+7i 3+4i 3+4i+21i+28i
= × = 2 2
3−4i 3+4i 3 +4

3+4i+21i−28 −25+25i
= 2 2
=
3 +4 25

= −1 + i

= −1 + i

Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

2 2 2
r (cos θ + sin θ) = 1 + 1

2 2 2
⇒ r (cos θ + sin θ) = 2

2 2 2
⇒ r = 2 [cos θ + sin θ = 1]

⇒ r = √2 [ Conventionally, r > 0]

∴ √2 cos θ = −1 and √2 sin θ = 1

∴: √2 cos θ = −1 and √2 sin θ = 1

−1 1
⇒ cos θ = and sin θ =
√2 √2

π 3π
∴ θ = π − =
4 4

: z = r cos θ + ir sin θ

3π 3π 3π 3π
= √2 cos + i√2 sin = √2 (cos + i sin )
4 4 4 4

This is the required polar form.
1+3i
(ii) Here z =
1−2i

1 + 3i 1 + 2i
= ×
1 − 2i 1 + 2i

1 + 2i + 3i − 6
=
1 + 4

−5 + 5i
= = −1 + i
5

Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

2 2 2
r (cos θ + sin θ) = 1 + 1

2 2 2
⇒ r (cos θ + sin θ) = 2

2 2 2
⇒ r = 2 [cos θ + sin θ = 1]

⇒ r = √2 [ Conventionally, r > 0]

∴ √2 cos θ = −1 and √2 sin θ = 1

−1 1
⇒ cos θ = and sin θ =
√2 √2

π 3π
∴ θ = π − =
4 4

∴ z = r cos θ + ir sin θ

3π 3π 3π 3π
= √2 cos + i√2 sin = √2 (cos + i sin )
4 4 4 4

This is the required polar form.

Page : 112 , Block Name : Miscellaneous Exercise

Q6 Solve the equation 3x 2 20
− 4x + = 0
3

The given quadratic equation is

2
This equation can also be written as 9x − 12x + 20 = 0
2
On comparing this equation with ax + bx + c = 0, we obtain

a = 9, b = −12, and c = 20

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = (−12) − 4 × 9 × 20 = 144 − 720 = −576

Therefore, the required solutions are

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = (−12) − 4 × 9 × 20 = 144 − 720 = −576

Therefore, the required solutions are

−b±√D −(−12)±√−576 12±√576i
= =
2a 2×9 18

12±24i 6(2±4i) 2±4i 2 4
= = = = ± i
18 18 3 3 3

Page : 112 , Block Name : Miscellaneous Exercise

Q7 Solve the equation x 2
− 2x +
3

2
= 0

2 3
x − 2x + = 0
2

2
This equation can also be written as 2x − 4x + 3 = 0

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

a = 2, b = −4, and c = 3

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = (−4) − 4 × 2 × 3 = 16 − 24 = −8

Therefore, the required solutions are

−b±√D −(−4)±√−8 4±2√2i
= =
2a 2×2 4
[√−1 = i]
2±√2i √2
= = 1 ± i
2 2

Page : 112 , Block Name : Miscellaneous Exercise

Q8 Solve the equation 27x 2
− 10x + 1 = 0

2
The given quadratic equation is 27x − 10x + 1 = 0

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

a = 27, b = −10, and c = 1

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = (−10) − 4 × 27 × 1 = 100 − 108 = −8

Therefore, the required solutions are

−b±√D −(−10)±√−8 10±2√2i
= =
2a 2×27 54

5±√2i 5 √2
= = ± i
27 27 27

Page : 112 , Block Name : Miscellaneous Exercise

Q9 Solve the equation 21x 2
− 28x + 10 = 0

2
The given quadratic equation is 21x − 28x + 10 = 0

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

a = 21, b = −28, and c = 10

Therefore, the discriminant of the given equation is

2 2
D = b − 4ac = (−28) − 4 × 21 × 10 = 784 − 840 = −56

Therefore, the required solutions are

−b±√D −(−28)±√−56 28±√56i
= =
2a 2×21 42

28±2√14i 28 2√14 2 √14
= = ± i = ± i
42 42 42 3 21

Page : 113 , Block Name : Miscellaneous Exercise

Q10 if z
z +z +1
∣ 1 2
∣
= 2 − i, z2 = 1 + i, find ∣
z −z +1 ∣
1
1 2

z1 = 2 − i, z2 = 1 + i

z +z +1 |(2−i)+(1+i)+1
∣ 1 2
∣
∴ =
∣ z1 −z2 +1 ∣ (2−i)−(1+i)+1

4 4
= = |
2−2i 2(1−i)

2(1+i)
∣ = ∣ ∣
2 1+i
= ∣
∣ ×
1−i 1+i ∣ ∣ 2
l −i
2
∣

∣ 2(1+i) ∣ 2
= [i = −1]
∣ 1+1 ∣

∣ 2(1 + i) ∣
= ∣ ∣
∣ 2 ∣

2 2
= |1 + i| = √1 + 1 = √2
z1 +z2 +1
∣ ∣ √2
∣ z −z +1 ∣ is
1 2

Page : 113 , Block Name : Miscellaneous Exercise

2
2 2
(x +1)
Q11 if a + ib =
(x+i)
2 2
, prove that a + b =
2 2
2x +1 (2x2 +1)

2
(x + i)
a + ib =
2
2x + 1
2 2
x + i + 2xi
=
2
2x + 1
2
x − 1 + i2x
=
2
2x + 1
2
x − 1 2x
= + i( )
2 2
2x + 1 2x + 1
2 2
2
x − 1 2x
2 2
∴ a + b = ( ) + ( )
2 2
2x + 1 2x + 1

4 2 2
x + 1 − 2x + 4x
=
2
(2x + 1)

4 2
x + 1 + 2x
=
2
2
(2x + 1)

z 2
(x + 1)
=
2
2
(2x + 1)
2
2
(x +1)
2 2
∴ a + b =
2
(2x2 +1)

Hence, proved.

Page : 113 , Block Name : Miscellaneous Exercise

Q12
Let z1 = 2 − i, z2 = −2 + i. Find
1
z1 z2 (ii) Im( )
¯
¯¯
(i) Re( ), z1 z 1
¯
¯¯
z1

z1 = 2 − i, z2 = −2 + i

2
Z1 I2 = (2 − i)(−2 + i) = −4 + 2i + 2i − i = −4 + 4i − (−1) = −3 + 4i
¯
¯¯¯
Z1 = 2 + i

Z1 Z2 −3+4i
∴ =
¯
¯¯¯ 2+i
Z1

2
z1 z2 (−3 + 4i)(2 − i) −6 + 3i + 8i − 4i −6 + 11i − 4(−1)
= = =
2 2 2 2
¯
¯¯
z1 (2 + i)(2 − i) 2 + l 2 + 1

−2 + 1li −2 11
= = + i
5 5 5

On comparing real parts, we obtain

z1 z2 −2
Re( ) =
¯
¯¯
z1 5

1 1 1 1
= = 2 2
=
z1 ¯
¯
z¯ (2−i)(2+i) (2) +(1) 5
1

On comparing imaginary parts, we obtain

1
lm( ) = 0
z1 ¯
¯
z¯
1

Page : 113 , Block Name : Miscellaneous Exercise

Q13 Find the modulus and argument of the complex number
1+2i

1−3i

1+2i
z = ,
1−3i
2 1+5i+6(−1)
1+2i 1+3i 1+3i+2i+6i
z = × = 2 2
=
1−3i 1+3i 1 +3 1+9

−5+5i −5 5i −1 1
= = + = + i
10 10 10 2 2

Let z = r cos θ + ir sin θ

−1 1
i.e., r cos θ = and r sin θ =
2 2
−1 1
i.e., r cos θ = and r sin θ =
2 2

On squaring and adding, we obtain

2 2
2 2 2 −1 1
r (cos θ + sin θ) = ( ) + ( )
2 2

2 1 1 1
⇒ r = + =
4 4 2
1
⇒ r = [ Conventionally, r > 0]
√2

1 −1 1 1
∴ cos θ = and sin θ =
√2 2 √2 2

−1 1
⇒ cos θ = and sin θ =
√2 √2

π 3π
∴ θ = π − = [ As θ lies in the II quadrant ]
4 4

3π
Therefore, the modulus and argument of the given complex number are √2 and
4

respectively.

Page : 113 , Block Name : Miscellaneous Exercise

Q14 Find the real number x and y if (x − iy)(3 + 5i) is the conjugate of − 6 − 24i

Let z = (x − ty)(3 + 5i)

2
z = 3x + 5xi − 3yi − 5yi = 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

5x − 3y = 24

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 − 15

⇒ y = −3

Thus, the values of x and y are 3 and − 3 respectlvely.

Page : 113 , Block Name : Miscellaneous Exercise

Q15 Find the modulas of
1+i 1−i
−
1−i 1+i

2 2
1 + i 1 − i (1 + i) − (1 − j)
− =
1 − i 1 + i (1 − i)(1 + i)

2 2
1 + i + 2i − 1 − i + 2i
=
2 2
l + 1

4i
∴ = = 2i
2

∣ 1 + i 1 − i ∣
∴ ∣ − √22 = 2
∣ = |2i| =
∣ 1 − i 1 + i ∣

Page : 113 , Block Name : Miscellaneous Exercise

Q16
3 u v 2 2
(x + iy) = u + iv, then show that + = 4 (x − y )
x y

3
(x + iy) = u + iv

3 3
⇒ x + (iy) + 3 ⋅ x ⋅ iy(x + iy) = u + iv

3 3 3 2 2
⇒ x + i y + 3x yi + 3xy i = u + iv

3 3 2 2
⇒ x − 4y + 3x yi − 3xy = u + iv

3 2 z 3
⇒ (x − 3xy ) + i (3x y − y ) = u + iv

3 2 2 3
u = x − 3xy , v = 3x y − y
3 2 2 3
u v x − 3xy 3x y − y
∴ + = +
x y x y

2 2 2 2
x (x − 3y ) y (3x − y )
= +
x y

2 2 2 2
= x − 3y + 3x − y

2 2
= 4 (x − 4y )

2 2
= 4 (x − 4y )

u v
2 2
∴ + = 4 (x − y )
x y

Page : 113 , Block Name : Miscellaneous Exercise

Q17 if α and β are different complex numbers with |β| = 1, then find ∣∣
β−α ∣
1−¯
¯¯
αβ¯ ∣

Let a = a + ib and β = x + iy

It is given that, |β| = 1

2 2
∴ √x + y = 1

2 2
⇒ x + y = I

∣ β − α ∣ ∣ (x + iy) − (a + ib) ∣
∣ ∣ = ∣ ∣
¯
¯¯¯
∣ 1 − αβ ∣ ∣ 1 − (a − ib)(x + iy) ∣

∣ (x − a) + i(y − b) ∣
= ∣ ∣
∣ 1 − (ax + aiy − ibx + by) ∣

(x − a) + i(y − b)
= |
(1 − ax − by) + i(bx − ay)
|(x−a)+i(y−b)|
=
(1−ax−by)+i(bx−ay)|

√(x−a)2 +(y−b)2

=
√(1−ax−by)2 +(bx−ay)2

√x2 +a2 −2ax+y 2 +b2 −2by

=
√1+a2 x2 +b2 y 2 −2ax+2abxy−2by+b2 x2 +a2 y 2 −2abxy

√(x2 +y 2 )+a2 +b2 −2ax−2by

=
√1+a2 (x2 +y 2 )+b2 (y 2 +x2 )−2ax−2by

√1+a2 +b2 −2ax−2by

= [U sin g(1)]
√1+a2 +b2 −2ax−2by

= 1

∣ β−α ∣
∴ = 1
∣ 1−¯αβ
¯
¯¯ ∣

Page : 113 , Block Name : Miscellaneous Exercise

Q18 Find the number of non-zero integral solutions of the equation |1 − i|
x x
= 2

x x
|1 − i| = 2
x

2 2 x
⇒ (√1 + (−1) ) = 2

x x
⇒ (√2) = 2
x
x
⇒ 22 = 2

⇒ x = 2x

⇒ 2x − x = 0

⇒ x = 0

Thus, 0 is the only integral solution of the given equation. Therefore, the nurnber of non-
integral solutions of Y--,e given equation is 0.

Page : 113 , Block Name : Miscellaneous Exercise

Q19
If (a + ib)(c + id)(e + if )(g + ih) = A + iB, then show that

2 2 2 2 2 2 2 2 2 2
(a + b ) (c + d ) (e + f ) (g + h ) = A + B

(a + ib)(c + id)(e + if )(g + ih) = A + iB

∴ (a + ib)(c + id)(e + if )(g + ih)| = |A + iB|

⇒ √a × √c = √A
2 2 2 2 2 2 2 2 2 2
+ b + d × √e + f × √g + h + B

On squaring both sides, we obtain

2 2 2 2 2 2 2 2 2 2
(a + b ) (c + c ) (e + f ) (g + h ) = A + B

Hence, proved.

Page : 113 , Block Name : Miscellaneous Exercise

Q20
m
1+i
( ) = 1, then find the least positive integral value of m
1−i

m
1+i
( ) = 1
1−i

m
1+i 1+i
⇒ ( × ) = 1
1−i 1+i

2 ′′
(1+i)
⇒ ( 2
) = 1
2
l +1

2 2
n
1 +i +2i
⇒ ( ) = 1
2

n
1−1+2i
⇒ ( ) = 1
2

n
2i
⇒ ( ) = 1
2

n
⇒ i = 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) .

Page : 113 , Block Name : Miscellaneous Exercise