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Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.4

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  • Last Updated : 09 Jul, 2021

Question 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 120o – i cos 120o

(viii) –16/(1 + i√3)

The polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) where,

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

Argument of complex number, θ = arg (Z) = tan–1 (y/x)

(i) 1 + i

Solution:

We are given, Z = 1 + i, so x = 1 and y = 1.

|Z| = √(12 + 12) = √2

θ = tan-1 (1/1) = tan-1 1

As x > 0 and y > 0, Z lies in 1st quadrant and the value of θ is 0 ≤ θ ≤ π/2.

So, θ = π/4 and Z = √2 (cos (π/4) + i sin (π/4)) 

Therefore, the polar form of (1 + i) is √2 (cos (π/4) + i sin (π/4)).

(ii) √3 + i

Solution:

We are given, Z = √3 + i, so x = √3 and y = 1.

|Z| = √((√3)2 + 12) = 2

θ = tan-1 (1/√3) 

As x > 0 and y > 0, Z lies in 1st quadrant and the value of θ is 0 ≤ θ ≤ π/2.

So, θ = π/6 and Z = 2 (cos (π/6) + i sin (π/6))

Therefore, the polar form of (√3 + i) is √2 (cos (π/6) + i sin (π/6)).

(iii) 1 – i

Solution:

We are given, Z = 1 – i, so x = 1 and y = –1.

|Z| = √(1)2 + (–1)2) = √2

θ = tan-1 (1/1) = tan-1 1  

As x > 0 and y < 0, Z lies in 4th quadrant and the value of θ is –π/2 ≤ θ ≤ 0.

So, θ = – π/4 and,

Z = √2 (cos (–π/4) + i sin (–π/4))

= √2 (cos (π/4) – i sin (π/4))

Therefore, the polar form of (1 – i) is √2 (cos (π/4) – i sin (π/4)).

(iv) (1 – i)/(1 + i)

Solution:

We are given, Z = (1 – i)/(1 + i).

Multiplying and dividing by (1 – i), we get,

Z = \frac{1 - i}{1 + i}×\frac{1 - i}{1 - i}

\frac{1+i^2-2i}{1-i^2}

\frac{1+(-1)-2i}{1-(-1)}

\frac{-2i}{2}

= 0 – i

So x = 0, y = –1 and |Z| = √(02 + (–1)2) = 1

θ =  tan-1 (1/0)

As x ≥ 0 and y < 0, Z lies in 4th quadrant and the value of θ is –π/2 ≤ θ ≤ 0.

So, θ = –π/2 and,

Z = 1 (cos (–π/2) + i sin (–π/2))

= cos (π/2) – i sin (π/2)

Therefore, the polar form of (1 – i)/(1 + i) is cos (π/2) – i sin (π/2).

(v) 1/(1 + i)

Solution:

We are given, Z = (1 – i)/(1 + i).

Multiplying and dividing by (1 – i), we get,

Z = \frac{1}{1 + i}×\frac{1 - i}{1 - i}

\frac{1-i}{1-i^2}

\frac{1-i}{1-(-1)}

\frac{1-i}{2}

= 1/2 – i/2

So x = 1/2, y = –1/2 and |Z| = √((1/2)2 + (–1/2)2) = √(2/4) = 1/√2

θ =  tan-1 ((1/2)/(1/2)) = tan–1 1

As x > 0 and y < 0, Z lies in 4th quadrant and the value of θ is –π/2 ≤ θ ≤ 0.

So, θ = –π/4 and,

Z = 1/√2 (cos (-π/4) + i sin (-π/4))

= 1/√2 (cos (π/4) – i sin (π/4))

Therefore, the polar form of 1/(1 + i) is 1/√2 (cos (π/4) – i sin (π/4)).

(vi) (1 + 2i)/(1 – 3i)

Solution:

We are given, Z = (1 + 2i)/(1 – 3i).

Multiplying and dividing by (1 + 3i), we get,

Z = \frac{1+2i}{1-3i}×\frac{1+3i}{1+3i}

\frac{1+3i+2i+6i^2}{1-9i^2}

\frac{1+5i+6(-1)}{1-9(-1)}

\frac{-5+5i}{10}

= –1/2 + i/2

So x = –1/2, y = 1/2 and |Z| = √((–1/2)2 + (1/2)2) = √(2/4) = 1/√2

θ =  tan-1 ((1/2)/(1/2)) = tan–1 1

As  x < 0 and y > 0, Z lies in 2nd quadrant and the value of θ is π/2 ≤ θ ≤ π.

So, θ = 3π/4 and Z = 1/√2 (cos (3π/4) + i sin (3π/4))

Therefore, the polar form of (1 + 2i)/(1 – 3i) is 1/√2 (cos (3π/4) + i sin (3π/4)).

(vii) sin 120o – i cos 120o

Solution:

We are given, Z = sin 120o – i cos 120o

= √3/2 – i (–1/2)

= √3/2 + i (1/2)

So x = √3/2, y = 1/2 and |Z| = √((√3/2)2 + (1/2)2) = √(3/4 + 1/4) = 1

θ = tan-1 ((1/2)/(√3/2)) = tan-1 (1/√3)

As x > 0 and y > 0, Z lies in 1st quadrant and the value of θ is 0 ≤ θ ≤ π/2.

So, θ = π/6 and Z = 1 (cos (π/6) + i sin (π/6))

Therefore, the polar form of √3/2 + i (1/2) is 1 (cos (π/6) + i sin (π/6)).

(viii) -16/(1 + i√3)

We are given, Z = –16/(1 + i√3).

Multiplying and dividing by (1 – i√3), we get,

Z = \frac{-16}{1+\sqrt{3}i}×\frac{1-\sqrt{3}i}{1-\sqrt{3}i}

\frac{-16+16\sqrt{3}i}{1-3i^2}

\frac{-16+16\sqrt{3}i}{4}         

= –4 + 4√3 i

So x = –4, y = 4/√3 and |Z| = √((–4)2 + (4√3)2) = √(16 + 48) = 8

θ = tan-1 (4√3/4) = tan-1 (√3)

As x < 0 and y > 0, Z lies in 2nd quadrant and the value of θ is π/2 ≤ θ ≤ π.

So, θ = 2π/3 and Z = 8 (cos (2π/3) + i sin (2π/3))

Therefore, the polar form of -16 / (1 + i√3) is 8 (cos (2π/3) + i sin (2π/3)).

Question 2. Write (i25)3 in polar form.

Solution:

We are given, 

 Z = (i25)3

= i75

= (i2)37. i

= (–1)37. i

= – i

= 0 – i

So x = 0, y = –1 and,

|Z| = √(x2 + y2)

= √(02 + (–1)2)

= 1

θ = tan–1 (|y| / |x|)

= tan–1 (1 / 0)

Since x ≥ 0 and y < 0, Z lies in 4th quadrant and the value of θ is π/2 ≤ θ ≤ 0. So, θ = –π/2.

Z = 1 (cos (–π/2) + i sin (–π/2))

= 1 (cos (π/2) – i sin (π/2))

Therefore, the polar form of (i25)3 is 1 (cos (π/2) – i sin (π/2)).

Question 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) \frac{1 - i}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}}

Solution:

(i) 1 + i tan α

We are given 1 + i tan α, so x =1 and y = tan α. 

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].

Case 1: If α ∈ [0, π/2)

|Z| = r = √(12 + tan2 α)

= √( sec2 α)

= sec α

θ = tan-1 (tan α/1)

= tan-1 (tan α)

= α 

So, Z = sec α (cos α + i sin α)

Therefore, polar form is sec α (cos α + i sin α).

Case 2: α ∈ (π/2, π]

|Z| = r = √(12 + tan2 α)

= √( sec2 α)

= – sec α

θ = tan-1 (tan α/1)

= tan-1 (tan α)

= –π + α

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

Therefore, polar form is –sec α (cos (α – π) + i sin (α – π)).

(ii) tan α – i

We are given tan α – i, so x =tan α and y = –1.

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].

Case 1: If α ∈ [0, π/2)

|Z| = r = √(tan2 α + 1)

= √( sec2 α)

= sec α

θ = tan-1 (1/tan α)

= tan-1 (cot α)

= α – π/2

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

Therefore, polar form is sec α (cos (α – π/2) + i sin (α – π/2)).

Case 2: α ∈ (π/2, π]

|Z| = r = √(tan2 α + 1)

= √( sec2 α)

= – sec α

θ = tan-1 (1/tan α)

= tan-1 (cot α)

= π/2 + α

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

Therefore, polar form is –sec α (cos (π/2 + α) + i sin (π/2 + α)).

(iii) 1 − sin α + i cos α

Let z = 1 − sin α + i cos α

As sine and cosine functions are periodic functions with period 2π, let us take α in [0, 2π].

Now, z = 1 − sin α + i cos α

\left| z \right| = \sqrt{\left( 1 - \sin\alpha \right)^2 + \cos^2 \alpha} = \sqrt{2 - \sin\alpha} = \sqrt{2}\sqrt{1 - \sin\alpha}

\left| z \right| = \sqrt{2}\sqrt{\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)^2} = \sqrt{2}\left| \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right|

Let θ be an acute angle given by,

tan θ = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|}

tan θ = \frac{\left| \cos\alpha \right|}{\left| 1 - \sin\alpha \right|} = \left| \frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{\left( \cos\frac{\alpha}{2} - \sin\frac{\alpha}{2} \right)^2} \right| = \left| \frac{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2} - \sin\frac{\alpha}{2}} \right|

tan θ = \left| \frac{1 + \tan\frac{\alpha}{2}}{1 - \tan\frac{\alpha}{2}} \right| = \left| \tan\left( \frac{\pi}{4} + \frac{\alpha}{2} \right) \right|

Case 1: When 0 ≤ α < π/2

In this case, we have,

|z| = √2(cos α/2 – sin α/2)

Also,

tan θ = |tan (π/4 + α/2)| = tan (π/4 + α/2)

θ = π/4 + α/2

Clearly, z lies in the first quadrant . Therefore, arg(z) = π/4 + α/2

Therefore, the polar form of z is √2(cos α/2 – sin α/2){cos (π/4 + α/2) + i sin (π/4 + α/2)}.

Case 2: When π/2 < α < 3π/2

In this case, we have,

|z| = |√2(cos α/2 – sin α/2)| = -√2(cos α/2 – sin α/2)

And, tan θ = |tan (π/4 + α/2)| = -tan (π/4 + α/2) = tan {π – (π/4 + α/2)} = tan (α/2 – 3π/4)   

θ = 3π/4 + α/2

Clearly, z lies in the fourth quadrant . Therefore, arg(z) = -θ = 3π/4 + α/2 = α/2 – 3π/4

Therefore, the polar form of z is -√2(cos α/2 – sin α/2){cos (α/2 – sin 3π/4) + i sin (α/2 – sin 3π/4)}..

Case 3: When 3π/2 < α < 2π

In this case, we have, 

|z| = |√2(cos α/2 – sin α/2)| = -√2(cos α/2 – sin α/2)

And tan θ = |tan (π/4 + α/2)| = tan (π/4 + α/2) = – tan {π – (π/4 + α/2)} = tan (α/2 – 3π/4)   

θ = α/2 – 3π/4

Clearly, z lies in the first quadrant . Therefore, arg(z) = θ = α/2 – 3π/4

Therefore, the polar form of z is -√2(cos α/2 – sin α/2){cos (α/2 – sin 3π/4) + i sin (α/2 – sin 3π/4)}.

(iv) \frac{1 - i}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}}

Let z = \frac{1 - i}{cos\frac{\pi}{3} + i sin\frac{\pi}{3}}

\frac{1 - i}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}

\frac{2 - 2i}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}}

\frac{2 - 2i - 2\sqrt{3}i + 2\sqrt{3} i^2}{1 + 3}

\frac{2 - 2\sqrt{3} - 2i(1 + \sqrt{3})}{4}

\frac{\left( 1 - \sqrt{3} \right) + i( - 1 - \sqrt{3})}{2}

\frac{\left( 1 - \sqrt{3} \right)}{2} + i\frac{( - 1 - \sqrt{3})}{2}

Now, z = \frac{\left( 1 - \sqrt{3} \right)}{2} + i\frac{( - 1 - \sqrt{3})}{2}

\left| z \right| = \sqrt{\left( \frac{1 - \sqrt{3}}{2} \right)^2 + \left( \frac{- 1 - \sqrt{3}}{2} \right)^2}

\sqrt{\left( \frac{1 + 3 - 2\sqrt{3}}{4} \right) + \left( \frac{1 + 3 + 2\sqrt{3}}{4} \right)}

\sqrt{\frac{8}{4}}

= √2

Let θ be an acute angle given by tan θ = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|}

tan θ = \frac{\left| \frac{1 + \sqrt{3}}{2} \right|}{\left| \frac{1 - \sqrt{3}}{2} \right|} = \left| \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \right| = \left| \frac{\tan\frac{\pi}{4} + \tan\frac{\pi}{3}}{1 - \tan\frac{\pi}{4}\tan\frac{\pi}{3}} \right|

tan θ = |tan (π/4 + π/3)| = |tan 7π/12| 

θ = 7π/12

Clearly, z lies in the fourth quadrant . Therefore, arg(z) = -7π/12

Therefore, the polar form of z is √2(cos 7π/12 – sin 7π/12).

Question 4. If z1 and z2 are two complex number such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that z_1 = -\bar{z_2}    .

Solution:

We are given |z1| = |z2| and arg (z1) + arg (z2) = π. Suppose arg (z1) = θ, then arg (z2) = π – θ.

We know z = |z| (cos θ + i sin θ)

z2 = |z2| (cos (π – θ) + i sin (π – θ))

= |z2| (–cos θ + i sin θ)

= – |z2| (cos θ – i sin θ)

The conjugate of z2\bar{z_2}        = – |z2| (cos θ + i sin θ)

Now L.H.S. = z1 = |z1| (cos θ + i sin θ)

= |z2| (cos θ + i sin θ)

= – [– |z2| (cos θ + i sin θ)]

-\bar{z_2}

= R.H.S.

Hence proved.

Question 5. If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, prove that arg (z1/z4) + arg (z2/z3) = 0

Solution:

We are given,

z_1=\bar{z_2}

z_3=\bar{z_4}

L.H.S. = arg (z1/z4) + arg (z2/z3)

= arg (z1) − arg (z4) + arg (z2) − arg (z3)

= [arg (z1) + arg (z2)] − [arg (z3) + arg (z4)] 

[arg(z_2)+arg(\bar{z_2})]−[arg(z_4)+arg(\bar{z_4})]

= 0 − 0

= 0 

= R.H.S

Hence proved.

Question 6. Express sin π/5 + i (1 – cos π/5) in polar form.

Solution:

We are given,

Z = sin π/5 + i (1 – cos π/5)

= 2 sin π/10 cos π/10 + i (2 sin2 π/10)

= 2 sin π/10 (cos π/10 + i sin π/10)

We know the polar form is given by r (cos θ + i sin θ).

Therefore, the polar form of the given expression is 2 sin π/10 (cos π/10 + i sin π/10).


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