RD Sharma Class 11 Solutions Chapter 13 Exercise 13.4 (Updated for 2024)

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RD Sharma Solutions Class 11 Maths Chapter 13 Exercise 13.4

RD Sharma Solutions for Class 11 Maths Chapter 13 Exercise 13.4

1. Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:

(i) 1 + i

(ii) √3 + i

(iii) 1 – i

(iv) (1 – i) / (1 + i)

(v) 1/(1 + i)

(vi) (1 + 2i) / (1 – 3i)

(vii) sin 120^o – i cos 120^o

(viii) -16 / (1 + i√3)

Solution:

We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)

Where,

|Z| = modulus of complex number = √(x² + y²)

θ = arg (z) = argument of complex number = tan^-1 (|y| / |x|)

(i) 1 + i

Given: Z = 1 + i

So now,

|Z| = √(x² + y²)

= √(1² + 1²)

= √(1 + 1)

= √2

θ = tan^-1 (|y| / |x|)

= tan^-1 (1 / 1)

= tan^-1 1

Since x > 0, y > 0 complex number lies in 1^st quadrant and the value of θ is 0⁰≤θ≤90⁰.

θ = π/4

Z = √2 (cos (π/4) + i sin (π/4))

∴ Polar form of (1 + i) is √2 (cos (π/4) + i sin (π/4))

(ii) √3 + i

Given: Z = √3 + i

So now,

|Z| = √(x² + y²)

= √((√3)² + 1²)

= √(3 + 1)

= √4

= 2

θ = tan^-1 (|y| / |x|)

= tan^-1 (1 / √3)

Since x > 0, y > 0 complex number lies in 1^st quadrant and the value of θ is 0⁰≤θ≤90⁰.

θ = π/6

Z = 2 (cos (π/6) + i sin (π/6))

∴ Polar form of (√3 + i) is 2 (cos (π/6) + i sin (π/6))

(iii) 1 – i

Given: Z = 1 – i

So now,

|Z| = √(x² + y²)

= √(1² + (-1)²)

= √(1 + 1)

= √2

θ = tan^-1 (|y| / |x|)

= tan^-1 (1 / 1)

= tan^-1 1

Since x > 0, y < 0 complex number lies in 4^th quadrant and the value of θ is -90⁰≤θ≤0⁰.

θ = -π/4

Z = √2 (cos (-π/4) + i sin (-π/4))

= √2 (cos (π/4) – i sin (π/4))

∴ The polar form of (1 – i) is √2 (cos (π/4) – i sin (π/4))

(iv) (1 – i) / (1 + i)

Given: Z = (1 – i) / (1 + i)

Let us multiply and divide by (1 – i), we get

= 0 – i

So now,

|Z| = √(x² + y²)

= √(0² + (-1)²)

= √(0 + 1)

= √1

θ = tan^-1 (|y| / |x|)

= tan^-1 (1 / 0)

= tan^-1 ∞

Since x ≥ 0, y < 0 complex number lies in 4^th quadrant and the value of θ is -90⁰≤θ≤0⁰.

θ = -π/2

Z = 1 (cos (-π/2) + i sin (-π/2))

= 1 (cos (π/2) – i sin (π/2))

∴ The polar form of (1 – i) / (1 + i) is 1 (cos (π/2) – i sin (π/2))

(v) 1/(1 + i)

Given: Z = 1 / (1 + i)

Let us multiply and divide by (1 – i), we get

So now,

|Z| = √(x² + y²)

= √((1/2)² + (-1/2)²)

= √(1/4 + 1/4)

= √(2/4)

= 1/√2

θ = tan^-1 (|y| / |x|)

= tan^-1 ((1/2) / (1/2))

= tan^-1 1

Since x > 0, y < 0 complex number lies in 4^th quadrant and the value of θ is -90⁰≤θ≤0⁰.

θ = -π/4

Z = 1/√2 (cos (-π/4) + i sin (-π/4))

= 1/√2 (cos (π/4) – i sin (π/4))

∴ Polar form of 1/(1 + i) is 1/√2 (cos (π/4) – i sin (π/4))

(vi) (1 + 2i) / (1 – 3i)

Given: Z = (1 + 2i) / (1 – 3i)

Let us multiply and divide by (1 + 3i), we get

So now,

|Z| = √(x² + y²)

= √((-1/2)² + (1/2)²)

= √(1/4 + 1/4)

= √(2/4)

= 1/√2

θ = tan^-1 (|y| / |x|)

= tan^-1 ((1/2) / (1/2))

= tan^-1 1

Since x < 0, y > 0 complex number lies in 2^nd quadrant and the value of θ is 90⁰≤θ≤180⁰.

θ = 3π/4

Z = 1/√2 (cos (3π/4) + i sin (3π/4))

∴ Polar form of (1 + 2i) / (1 – 3i) is 1/√2 (cos (3π/4) + i sin (3π/4))

(vii) sin 120^o – i cos 120^o

Given: Z = sin 120^o – i cos 120^o

= √3/2 – i (-1/2)

= √3/2 + i (1/2)

So now,

|Z| = √(x² + y²)

= √((√3/2)² + (1/2)²)

= √(3/4 + 1/4)

= √(4/4)

= √1

= 1

θ = tan^-1 (|y| / |x|)

= tan^-1 ((1/2) / (√3/2))

= tan^-1 (1/√3)

Since x > 0, y > 0 complex number lies in 1^st quadrant and the value of θ is 0⁰≤θ≤90⁰.

θ = π/6

Z = 1 (cos (π/6) + i sin (π/6))

∴ Polar form of √3/2 + i (1/2) is 1 (cos (π/6) + i sin (π/6))

(viii) -16 / (1 + i√3)

Given: Z = -16 / (1 + i√3)

Let us multiply and divide by (1 – i√3), we get

So now,

|Z| = √(x² + y²)

= √((-4)² + (4√3)²)

= √(16 + 48)

= √(64)

= 8

θ = tan^-1 (|y| / |x|)

= tan^-1 ((4√3) / 4)

= tan^-1 (√3)

Since x < 0, y > 0 complex number lies in 2^nd quadrant and the value of θ is 90⁰≤θ≤180⁰.

θ = 2π/3

Z = 8 (cos (2π/3) + i sin (2π/3))

∴ Polar form of -16 / (1 + i√3) is 8 (cos (2π/3) + i sin (2π/3))

2. Write (i²⁵)³ in polar form.

Solution:

Given: Z = (i²⁵)³

= i⁷⁵

= i⁷⁴. i

= (i²)³⁷. i

= (-1)³⁷. i

= (-1). i

= – i

= 0 – i

So now,

|Z| = √(x² + y²)

= √(0² + (-1)²)

= √(0 + 1)

= √1

θ = tan^-1 (|y| / |x|)

= tan^-1 (1 / 0)

= tan^-1 ∞

Since x ≥ 0, y < 0 complex number lies in 4^th quadrant and the value of θ is -90⁰≤θ≤0⁰.

θ = -π/2

Z = 1 (cos (-π/2) + i sin (-π/2))

= 1 (cos (π/2) – i sin (π/2))

∴ Polar form of (i²⁵)³ is 1 (cos (π/2) – i sin (π/2))

3. Express the following complex numbers in the form r (cos θ + i sin θ):

(i) 1 + i tan α

(ii) tan α – i

(iii) 1 – sin α + i cos α

(iv) (1 – i) / (cos π/3 + i sin π/3)

Solution:

(i) 1 + i tan α

Given: Z = 1 + i tan α

We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)

Where,

|Z| = modulus of complex number = √(x² + y²)

θ = arg (z) = argument of complex number = tan^-1 (|y| / |x|)

We also know that tan α is a periodic function with period π.

So α is lying in the interval [0, π/2) ∪ (π/2, π].

Let us consider case 1:

α ∈ [0, π/2)

So now,

|Z| = r = √(x² + y²)

= √(1² + tan² α)

= √( sec² α)

= |sec α| since, sec α is positive in the interval [0, π/2)

θ = tan^-1 (|y| / |x|)

= tan^-1 (tan α / 1)

= tan^-1 (tan α)

= α since, tan α is positive in the interval [0, π/2)

∴ Polar form is Z = sec α (cos α + i sin α)

Let us consider case 2:

α ∈ (π/2, π]

So now,

|Z| = r = √(x² + y²)

= √(1² + tan² α)

= √( sec² α)

= |sec α|

= – sec α since, sec α is negative in the interval (π/2, π]

θ = tan^-1 (|y| / |x|)

= tan^-1 (tan α / 1)

= tan^-1 (tan α)

= -π + α since, tan α is negative in the interval (π/2, π]

θ = -π + α [since, θ lies in 4^th quadrant]

Z = -sec α (cos (α – π) + i sin (α – π))

∴ Polar form is Z = -sec α (cos (α – π) + i sin (α – π))

(ii) tan α – i

Given: Z = tan α – i

We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)

Where,

|Z| = modulus of complex number = √(x² + y²)

θ = arg (z) = argument of complex number = tan^-1 (|y| / |x|)

We also know that tan α is a periodic function with period π.

So α is lying in the interval [0, π/2) ∪ (π/2, π].

Let us consider case 1:

α ∈ [0, π/2)

So now,

|Z| = r = √(x² + y²)

= √(tan² α + 1²)

= √( sec² α)

= |sec α| since, sec α is positive in the interval [0, π/2)

= sec α

θ = tan^-1 (|y| / |x|)

= tan^-1 (1/tan α)

= tan^-1 (cot α) since, cot α is positive in the interval [0, π/2)

= α – π/2 [since, θ lies in 4^th quadrant]

Z = sec α (cos (α – π/2) + i sin (α – π/2))

∴ Polar form is Z = sec α (cos (α – π/2) + i sin (α – π/2))

Let us consider case 2:

α ∈ (π/2, π]

So now,

|Z| = r = √(x² + y²)

= √(tan² α + 1²)

= √( sec² α)

= |sec α|

= – sec α since, sec α is negative in the interval (π/2, π]

θ = tan^-1 (|y| / |x|)

= tan^-1 (1/tan α)

= tan^-1 (cot α)

= π/2 + α since, cot α is negative in the interval (π/2, π]

θ = π/2 + α [since, θ lies in 3^th quadrant]

Z = -sec α (cos (π/2 + α) + i sin (π/2 + α))

∴ Polar form is Z = -sec α (cos (π/2 + α) + i sin (π/2 + α))

(iii) 1 – sin α + i cos α

Given: Z = 1 – sin α + i cos α

By using the formulas,

Sin² θ + cos² θ = 1

Sin 2θ = 2 sin θ cos θ

Cos 2θ = cos² θ – sin² θ

So,

z= (sin²(α/2) + cos²(α/2) – 2 sin(α/2) cos(α/2)) + i (cos²(α/2) – sin²(α/2))

= (cos(α/2) – sin(α/2))² + i (cos²(α/2) – sin²(α/2))

We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)

Where,

|Z| = modulus of complex number = √(x² + y²)

θ = arg (z) = argument of complex number = tan^-1 (|y| / |x|)

Now,

We know that sine and cosine functions are periodic with period 2π

Here we have 3 intervals:

0 ≤ α ≤ π/2

π/2 ≤ α ≤ 3π/2

3π/2 ≤ α ≤ 2π

Let us consider case 1:

In the interval 0 ≤ α ≤ π/2

Cos (α/2) > sin (α/2) and also 0 < π/4 + α/2 < π/2

So,

∴ Polar form is Z = √2 (cos (α/2) – sin (α/2)) (cos (π/4 + α/2) + i sin (π/4 + α/2))

Let us consider case 2:

In the interval π/2 ≤ α ≤ 3π/2

Cos (α/2) < sin (α/2) and also π/2 < π/4 + α/2 < π

So,

Since, (1 – sin α) > 0 and cos α < 0 [Z lies in 4^th quadrant]

= α/2 – 3π/4

∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 – 3π/4) + i sin (α/2 – 3π/4))

Let us consider case 3:

In the interval 3π/2 ≤ α ≤ 2π

Cos (α/2) < sin (α/2) and also π < π/4 + α/2 < 5π/4

So,

θ = tan^-1 (tan (π/4 + α/2))

= π – (π/4 + α/2) [since, θ lies in 1st quadrant and tan’s period is π]

= α/2 – 3π/4

∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 – 3π/4) + i sin (α/2 – 3π/4))

(iv) (1 – i) / (cos π/3 + i sin π/3)

Given: Z = (1 – i) / (cos π/3 + i sin π/3)

Let us multiply and divide by (1 – i√3), we get

We know that the polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)

Where,

|Z| = modulus of complex number = √(x² + y²)

θ = arg (z) = argument of complex number = tan^-1 (|y| / |x|)

Now,

Since x < 0, y < 0 complex number lies in 3^rd quadrant and the value of θ is 180⁰≤θ≤-90⁰.

= tan^-1 (2 + √3)

= -7π/12

Z = √2 (cos (-7π/12) + i sin (-7π/12))

= √2 (cos (7π/12) – i sin (7π/12))

∴ Polar form of (1 – i) / (cos π/3 + i sin π/3) is √2 (cos (7π/12) – i sin (7π/12))

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