Exercise 5.2 Solutions Complex Numbers and Quadratic Equations – NCERT Class 11 Mathematics Chapter 5

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

1. z = – 1 – i √3

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 – i √3 with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = – √3

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (-1)2 + (- √3)2 = 1 + 3 = 4

⇒ r2Cos2θ + r2Sin2θ = 4

r2(Cos2θ + Sin2θ) = 4

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 4

r2 = 4

r = ± 2
Thus r = 2 or -2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 2

Hence Modulus of Complex Number z = – 1 – i √3 is 2


Now let’s find out Argument of the Complex Number z = – 1 – i √3

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 – i √3 with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = – √3

Replacing r = 2 in these equations
2Cosθ = – 1
2Sinθ = – √3

⇒ Cosθ = – 1/2
⇒ Sinθ = – √3/2

⇒ θ = 4𝜋/3
As both Cosine and Sine are negative thus θ must lie in the third quadrant, but by definition Argument of a Complex Number lie in range − 𝜋 < 𝜃 ≤ 𝜋.

So θ if taken in reverse direction from fourth quadrant of x-y plane will be – (2𝜋 – 4𝜋/3) = – 2𝜋/3
Which does lie in range − 𝜋 < 𝜃 ≤ 𝜋.

Thus Argument of Complex Number z = – 1 – i √3 is – 2𝜋/3


2. z = – √3 + i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – √3 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – √3
rSinθ = 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (- √3)2 + (1)2 = 3 + 1 = 4

⇒ r2Cos2θ + r2Sin2θ = 4

r2(Cos2θ + Sin2θ) = 4

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 4

r2 = 4

r = ± 2
Thus r = 2 or -2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 2

Hence Modulus of Complex Number z = – √3 + i is 2


Now let’s find out Argument of the Complex Number z = – √3 + i
Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – √3 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – √3
rSinθ = 1

Replacing r = 2 in these equations
2Cosθ = – √3
2Sinθ = 1

⇒ Cosθ = – √3/2
⇒ Sinθ = 1/2

⇒ θ = 𝜋/6

As Cosine is negative but Sine is positive thus θ must lie in the second quadrant and as θ = 𝜋/6 lie in range − 𝜋 < 𝜃 ≤ 𝜋. Thus it is the argument of given Complex Number z = – √3 + i.

Thus Argument of Complex Number z = – √3 + i is 𝜋/6


Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = 1 – i with z = rCosθ + i(rSinθ)
We get
rCosθ = 1
rSinθ = – 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (1)2 + (- 1)2 = 1 + 1 = 2

⇒ r2Cos2θ + r2Sin2θ = 2

r2(Cos2θ + Sin2θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 2

r2 = 2

r = ± √2
Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number z = 1 – i is r = √2


Now let’s find out Argument of the Complex Number z = 1 – i
Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = 1 – i with z = rCosθ + i(rSinθ)
We get
rCosθ = 1
rSinθ = – 1

Replacing r = √2 in these equations
√2Cosθ = 1
√2Sinθ = – 1

⇒ Cosθ = 1/√2
⇒ Sinθ = – 1/√2

As Sine is negative but Cosine is positive thus θ must lie in fourth quadrant
So for Cosθ = 1/√2 and Sinθ = – 1/√2 value of θ = 7𝜋/4

But by definition argument of a Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋
And as θ = 7𝜋/4 lies in fourth quadrant thus it can also be written as
– (2𝜋 – 7𝜋/4) = – 𝜋/4

Thus θ = – 𝜋/4 which does lie in range − 𝜋 < 𝜃 ≤ 𝜋 and is thus Argument of Complex Number z = 1 – i

So
r = √2
θ = – 𝜋/4

Polar Form = rCosθ + i(rSinθ) = √2Cos(- 𝜋/4) + i(√2Sin(- 𝜋/4))
= √2[Cos(- 𝜋/4) + i Sin(- 𝜋/4)]

Thus Polar Form of Complex Number z = 1 – i is √2[Cos(- 𝜋/4) + i Sin(- 𝜋/4)]


4. – 1 + i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (-1)2 + (1)2 = 1 + 1 = 2

⇒ r2Cos2θ + r2Sin2θ = 2

r2(Cos2θ + Sin2θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 2

r2 = 2

r = ± √2
Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number z = – 1 + i is √2


Now let’s find out Argument of the Complex Number z = – 1 + i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = 1

Replacing r = √2 in these equations
√2Cosθ = – 1
√2Sinθ = 1

⇒ Cosθ = – 1/√2
⇒ Sinθ = 1/√2

As Cosine is negative but Sine is positive thus θ must lie in the second quadrant
⇒ θ = 3𝜋/4

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 3𝜋/4 does lies in.

Hence it’s the argument of given Complex Number z = – 1 + i

Thus
r = √2
θ = 3𝜋/4

Polar Form of Complex Number = rCosθ + i(rSinθ)
Replacing r and θ

⇒ Polar Form of Complex Number – 1 + i = √2Cos(3𝜋/4) + i[√2Sin(3𝜋/4)]
= √2[Cos(3𝜋/4) + iSin(3𝜋/4)]

Thus Polar Form of Complex Number z = – 1 + i is √2[Cos(3𝜋/4) + iSin(3𝜋/4)]


5. – 1 – i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 – i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = – 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (- 1)2 + (- 1)2 = 1 + 1 = 2

⇒ r2Cos2θ + r2Sin2θ = 2

r2(Cos2θ + Sin2θ) = 2

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 2

r2 = 2

r = ± √2
Thus r = √2 or – √2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = √2

Hence Modulus of Complex Number z = – 1 – i is √2


Now let’s find out Argument of the Complex Number z = – 1 – i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 1 – i with z = rCosθ + i(rSinθ)
We get
rCosθ = – 1
rSinθ = – 1

Replacing r = √2 in these equations
√2Cosθ = – 1
√2Sinθ = – 1

⇒ Cosθ = – 1/√2
⇒ Sinθ = – 1/√2

As Cosine is negative but Sine is positive thus θ must lie in the third quadrant
⇒ θ = 5𝜋/4

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 5𝜋/4 does not lies in.

Thus measuring θ = 5𝜋/4 from fourth quadrant in the reverse direction we get
– (2𝜋 – 5𝜋/4) = – 3𝜋/4 which does lies in range − 𝜋 < 𝜃 ≤ 𝜋

Hence Argument θ of Complex Number – 1 – i is -3𝜋/4

Thus
r = √2
θ = – 3𝜋/4

Polar Form of Complex Number = rCosθ + i(rSinθ)
Replacing r and θ

⇒ Polar Form of Complex Number – 1 – i = √2Cos(- 3𝜋/4) + i[√2Sin(- 3𝜋/4)]
= √2[Cos(- 3𝜋/4) + iSin( 3𝜋/4)]

Thus Polar Form of Complex Number z = – 1 – i is √2[Cos(- 3𝜋/4) + iSin(- 3𝜋/4)]


6. – 3

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 3 with z = rCosθ + i(rSinθ)
We get
rCosθ = – 3
rSinθ = 0

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (- 3)2 + (0)2 = 9 + 0 = 9

⇒ r2Cos2θ + r2Sin2θ = 9

r2(Cos2θ + Sin2θ) = 9

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 9

r2 = 9

r = ± 3
Thus r = 3 or 3 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 3

Hence Modulus of Complex Number z = – 3 is 3


Now let’s find out Argument of the Complex Number z = – 3

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = – 3 with z = rCosθ + i(rSinθ)
We get
rCosθ = – 3
rSinθ = 0

Replacing r = 3 in these equations
3Cosθ = – 3
3Sinθ = 0

⇒ Cosθ = – 3/3 = – 1
⇒ Sinθ = 0

As Cosine is negative but Sine is positive thus θ must lie in the second quadrant
⇒ θ = 𝜋

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 𝜋 does lies in.

Thus θ = 𝜋 is the Argument of given Complex Number z = – 3

Thus
r = 3
θ = 𝜋

Polar Form of Complex Number = rCosθ + i(rSinθ)
⇒ 3Cos𝜋 + i(3Sin𝜋)
= 3[Cos𝜋 + i Sin𝜋]

Hence Polar Form of Complex Number z = – 3 is 3[Cos𝜋 + i Sin𝜋]


7. √3 + i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = √3 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = √3
rSinθ = 1

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (√3)2 + (1)2 = 3 + 1 = 4

⇒ r2Cos2θ + r2Sin2θ = 4

r2(Cos2θ + Sin2θ) = 4

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 4

r2 = 4

r = ± 2
Thus r = 2 or – 2 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 2

Hence Modulus of Complex Number z = √3 + i is 2


Now let’s find out Argument of the Complex Number z = √3 + i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = √3 + i with z = rCosθ + i(rSinθ)
We get
rCosθ = √3
rSinθ = 1

Replacing r = 2 in these equations
2Cosθ = √3
2Sinθ = 1

⇒ Cosθ = √3/2
⇒ Sinθ = 1/2

As both of Sine and Cosine are positive thus θ must lie in first quadrant
⇒ θ = 𝜋/6

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 𝜋/6 does lies in.

Hence Argument of Complex Number z = √3 + i is θ = 𝜋/6

Thus
r = 2
θ = 𝜋/6

Polar Form of Complex Number = rCosθ + i(rSinθ)
= 2Cos(𝜋/6) + i(2Sin(𝜋/6))
= 2[Cos(𝜋/6) + i Sin(𝜋/6)]

Thus Polar form of Complex Number z = √3 + i is 2[Cos(𝜋/6) + i Sin(𝜋/6)]


8. i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = i with z = rCosθ + i(rSinθ)
We get
rCosθ = 0
rSinθ = i

Squaring both of these equations and then adding
r2Cos2θ + r2Sin2θ = (0)2 + (1)2 = 0 + 1 = 1

⇒ r2Cos2θ + r2Sin2θ = 1

r2(Cos2θ + Sin2θ) = 1

Sine square plus Cos square is always equal to 1 for any value of angle
⇒ r2(1) = 1

r2 = 1

r = ± 1
Thus r = 1 or – 1 but by definition of Modulus of a Complex Number it cannot be negative thus only possible value of r = 1

Hence Modulus of Complex Number z = i is 1


Now let’s find out Argument of the Complex Number z = i

Any Complex Number of form z = a + ib can be written as z = rCosθ + i(rSinθ) where
rCosθ = a
rSinθ = b

On Comparing z = i with z = rCosθ + i(rSinθ)
We get
rCosθ = 0
rSinθ = 1

Replacing r = 1 in these equations
1 × Cosθ = 0
1 × Sinθ = 1

⇒ Cosθ = 0
⇒ Sinθ = 1

As both of Sine and Cosine are positive thus θ must lie in first quadrant
⇒ θ = 𝜋/2

And by definition Argument of any Complex Number θ must lie in range − 𝜋 < 𝜃 ≤ 𝜋 which θ = 𝜋/2 does lies in.

Hence Argument of Complex Number z = i is θ = 𝜋/2

Thus
r = 1
θ = 𝜋/2

Polar Form of Complex Number = rCosθ + i(rSinθ)
= 1 × Cos(𝜋/2) + i(1 × Sin(𝜋/2))
= Cos(𝜋/2) + i Sin(𝜋/2)

Thus Polar form of Complex Number z = i is Cos(𝜋/2) + i Sin(𝜋/2)


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