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

• 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) âˆš3 + i

(iii) 1 â€“ i

(iv) (1 â€“ i)/(1 + i)

(v) 1/(1 + i)

(vi) (1 + 2i)/(1 â€“ 3i)

(vii) sin 120o â€“ i cos 120o

(viii) â€“16/(1 + iâˆš3)

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

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

Argument of complex number, Î¸ = arg (Z) = tanâ€“1 (y/x)

### (i) 1 + i

Solution:

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

|Z| = âˆš(12 + 12) = âˆš2

Î¸ = tan-1 (1/1) = tan-1 1

As x > 0 and y > 0, Z lies in 1st quadrant and the value of Î¸ is 0 â‰¤ Î¸ â‰¤ Ï€/2.

So, Î¸ = Ï€/4 and Z = âˆš2 (cos (Ï€/4) + i sin (Ï€/4))

Therefore, the polar form of (1 + i) is âˆš2 (cos (Ï€/4) + i sin (Ï€/4)).

### (ii) âˆš3 + i

Solution:

We are given, Z = âˆš3 + i, so x = âˆš3 and y = 1.

|Z| = âˆš((âˆš3)2 + 12) = 2

Î¸ = tan-1 (1/âˆš3)

As x > 0 and y > 0, Z lies in 1st quadrant and the value of Î¸ is 0 â‰¤ Î¸ â‰¤ Ï€/2.

So, Î¸ = Ï€/6 and Z = 2 (cos (Ï€/6) + i sin (Ï€/6))

Therefore, the polar form of (âˆš3 + i) is âˆš2 (cos (Ï€/6) + i sin (Ï€/6)).

### (iii) 1 â€“ i

Solution:

We are given, Z = 1 â€“ i, so x = 1 and y = â€“1.

|Z| = âˆš(1)2 + (â€“1)2) = âˆš2

Î¸ = tan-1 (1/1) = tan-1 1

As x > 0 and y < 0, Z lies in 4th quadrant and the value of Î¸ is â€“Ï€/2 â‰¤ Î¸ â‰¤ 0.

So, Î¸ = â€“ Ï€/4 and,

Z = âˆš2 (cos (â€“Ï€/4) + i sin (â€“Ï€/4))

= âˆš2 (cos (Ï€/4) â€“ i sin (Ï€/4))

Therefore, the polar form of (1 â€“ i) is âˆš2 (cos (Ï€/4) â€“ i sin (Ï€/4)).

### (iv) (1 â€“ i)/(1 + i)

Solution:

We are given, Z = (1 â€“ i)/(1 + i).

Multiplying and dividing by (1 â€“ i), we get,

Z =

= 0 â€“ i

So x = 0, y = â€“1 and |Z| = âˆš(02 + (â€“1)2) = 1

Î¸ =  tan-1 (1/0)

As x â‰¥ 0 and y < 0, Z lies in 4th quadrant and the value of Î¸ is â€“Ï€/2 â‰¤ Î¸ â‰¤ 0.

So, Î¸ = â€“Ï€/2 and,

Z = 1 (cos (â€“Ï€/2) + i sin (â€“Ï€/2))

= cos (Ï€/2) â€“ i sin (Ï€/2)

Therefore, the polar form of (1 â€“ i)/(1 + i) is cos (Ï€/2) â€“ i sin (Ï€/2).

### (v) 1/(1 + i)

Solution:

We are given, Z = (1 â€“ i)/(1 + i).

Multiplying and dividing by (1 â€“ i), we get,

Z =

= 1/2 â€“ i/2

So x = 1/2, y = â€“1/2 and |Z| = âˆš((1/2)2 + (â€“1/2)2) = âˆš(2/4) = 1/âˆš2

Î¸ =  tan-1 ((1/2)/(1/2)) = tanâ€“1 1

As x > 0 and y < 0, Z lies in 4th quadrant and the value of Î¸ is â€“Ï€/2 â‰¤ Î¸ â‰¤ 0.

So, Î¸ = â€“Ï€/4 and,

Z = 1/âˆš2 (cos (-Ï€/4) + i sin (-Ï€/4))

= 1/âˆš2 (cos (Ï€/4) â€“ i sin (Ï€/4))

Therefore, the polar form of 1/(1 + i) is 1/âˆš2 (cos (Ï€/4) â€“ i sin (Ï€/4)).

### (vi) (1 + 2i)/(1 â€“ 3i)

Solution:

We are given, Z = (1 + 2i)/(1 â€“ 3i).

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

Z =

= â€“1/2 + i/2

So x = â€“1/2, y = 1/2 and |Z| = âˆš((â€“1/2)2 + (1/2)2) = âˆš(2/4) = 1/âˆš2

Î¸ =  tan-1 ((1/2)/(1/2)) = tanâ€“1 1

As  x < 0 and y > 0, Z lies in 2nd quadrant and the value of Î¸ is Ï€/2 â‰¤ Î¸ â‰¤ Ï€.

So, Î¸ = 3Ï€/4 and Z = 1/âˆš2 (cos (3Ï€/4) + i sin (3Ï€/4))

Therefore, the polar form of (1 + 2i)/(1 â€“ 3i) is 1/âˆš2 (cos (3Ï€/4) + i sin (3Ï€/4)).

### (vii) sin 120o â€“ i cos 120o

Solution:

We are given, Z = sin 120o â€“ i cos 120o

= âˆš3/2 â€“ i (â€“1/2)

= âˆš3/2 + i (1/2)

So x = âˆš3/2, y = 1/2 and |Z| = âˆš((âˆš3/2)2 + (1/2)2) = âˆš(3/4 + 1/4) = 1

Î¸ = tan-1 ((1/2)/(âˆš3/2)) = tan-1 (1/âˆš3)

As x > 0 and y > 0, Z lies in 1st quadrant and the value of Î¸ is 0 â‰¤ Î¸ â‰¤ Ï€/2.

So, Î¸ = Ï€/6 and Z = 1 (cos (Ï€/6) + i sin (Ï€/6))

Therefore, the polar form of âˆš3/2 + i (1/2) is 1 (cos (Ï€/6) + i sin (Ï€/6)).

### (viii) -16/(1 + iâˆš3)

We are given, Z = â€“16/(1 + iâˆš3).

Multiplying and dividing by (1 â€“ iâˆš3), we get,

Z =

= â€“4 + 4âˆš3 i

So x = â€“4, y = 4/âˆš3 and |Z| = âˆš((â€“4)2 + (4âˆš3)2) = âˆš(16 + 48) = 8

Î¸ = tan-1 (4âˆš3/4) = tan-1 (âˆš3)

As x < 0 and y > 0, Z lies in 2nd quadrant and the value of Î¸ is Ï€/2 â‰¤ Î¸ â‰¤ Ï€.

So, Î¸ = 2Ï€/3 and Z = 8 (cos (2Ï€/3) + i sin (2Ï€/3))

Therefore, the polar form of -16 / (1 + iâˆš3) is 8 (cos (2Ï€/3) + i sin (2Ï€/3)).

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

Solution:

We are given,

Z = (i25)3

= i75

= (i2)37. i

= (â€“1)37. i

= â€“ i

= 0 â€“ i

So x = 0, y = â€“1 and,

|Z| = âˆš(x2 + y2)

= âˆš(02 + (â€“1)2)

= 1

Î¸ = tanâ€“1 (|y| / |x|)

= tanâ€“1 (1 / 0)

Since x â‰¥ 0 and y < 0, Z lies in 4th quadrant and the value of Î¸ is Ï€/2 â‰¤ Î¸ â‰¤ 0. So, Î¸ = â€“Ï€/2.

Z = 1 (cos (â€“Ï€/2) + i sin (â€“Ï€/2))

= 1 (cos (Ï€/2) â€“ i sin (Ï€/2))

Therefore, the polar form of (i25)3 is 1 (cos (Ï€/2) â€“ i sin (Ï€/2)).

### Question 3. Express the following complex numbers in the form r (cos Î¸ + i sin Î¸):

(i) 1 + i tan Î±

(ii) tan Î± â€“ i

(iii) 1 âˆ’ sin Î± + i cos Î±

(iv)

Solution:

(i) 1 + i tan Î±

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

We also know that tan Î± is a periodic function with period Ï€.

So Î± is lying in the interval [0, Ï€/2) âˆª (Ï€/2, Ï€].

Case 1: If Î± âˆˆ [0, Ï€/2)

|Z| = r = âˆš(12 + tan2 Î±)

= âˆš( sec2 Î±)

= sec Î±

Î¸ = tan-1 (tan Î±/1)

= tan-1 (tan Î±)

= Î±

So, Z = sec Î± (cos Î± + i sin Î±)

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

Case 2: Î± âˆˆ (Ï€/2, Ï€]

|Z| = r = âˆš(12 + tan2 Î±)

= âˆš( sec2 Î±)

= â€“ sec Î±

Î¸ = tan-1 (tan Î±/1)

= tan-1 (tan Î±)

= â€“Ï€ + Î±

So, Z = â€“sec Î± (cos (Î± â€“ Ï€) + i sin (Î± â€“ Ï€))

Therefore, polar form is â€“sec Î± (cos (Î± â€“ Ï€) + i sin (Î± â€“ Ï€)).

(ii) tan Î± â€“ i

We are given tan Î± â€“ i, so x =tan Î± and y = â€“1.

We also know that tan Î± is a periodic function with period Ï€.

So Î± is lying in the interval [0, Ï€/2) âˆª (Ï€/2, Ï€].

Case 1: If Î± âˆˆ [0, Ï€/2)

|Z| = r = âˆš(tan2 Î± + 1)

= âˆš( sec2 Î±)

= sec Î±

Î¸ = tan-1 (1/tan Î±)

= tan-1 (cot Î±)

= Î± â€“ Ï€/2

So, Z = sec Î± (cos (Î± â€“ Ï€/2) + i sin (Î± â€“ Ï€/2))

Therefore, polar form is sec Î± (cos (Î± â€“ Ï€/2) + i sin (Î± â€“ Ï€/2)).

Case 2: Î± âˆˆ (Ï€/2, Ï€]

|Z| = r = âˆš(tan2 Î± + 1)

= âˆš( sec2 Î±)

= â€“ sec Î±

Î¸ = tan-1 (1/tan Î±)

= tan-1 (cot Î±)

= Ï€/2 + Î±

So, Z =  â€“sec Î± (cos (Ï€/2 + Î±) + i sin (Ï€/2 + Î±))

Therefore, polar form is â€“sec Î± (cos (Ï€/2 + Î±) + i sin (Ï€/2 + Î±)).

(iii) 1 âˆ’ sin Î± + i cos Î±

Let z = 1 âˆ’ sin Î± + i cos Î±

As sine and cosine functions are periodic functions with period 2Ï€, let us take Î± in [0, 2Ï€].

Now, z = 1 âˆ’ sin Î± + i cos Î±

Let Î¸ be an acute angle given by,

tan Î¸ =

tan Î¸ =

tan Î¸ =

Case 1: When 0 â‰¤ Î± < Ï€/2

In this case, we have,

|z| = âˆš2(cos Î±/2 – sin Î±/2)

Also,

tan Î¸ = |tan (Ï€/4 + Î±/2)| = tan (Ï€/4 + Î±/2)

Î¸ = Ï€/4 + Î±/2

Clearly, z lies in the first quadrant . Therefore, arg(z) = Ï€/4 + Î±/2

Therefore, the polar form of z is âˆš2(cos Î±/2 – sin Î±/2){cos (Ï€/4 + Î±/2) + i sin (Ï€/4 + Î±/2)}.

Case 2: When Ï€/2 < Î± < 3Ï€/2

In this case, we have,

|z| = |âˆš2(cos Î±/2 – sin Î±/2)| = -âˆš2(cos Î±/2 – sin Î±/2)

And, tan Î¸ = |tan (Ï€/4 + Î±/2)| = -tan (Ï€/4 + Î±/2) = tan {Ï€ – (Ï€/4 + Î±/2)} = tan (Î±/2 – 3Ï€/4)

Î¸ = 3Ï€/4 + Î±/2

Clearly, z lies in the fourth quadrant . Therefore, arg(z) = -Î¸ = 3Ï€/4 + Î±/2 = Î±/2 – 3Ï€/4

Therefore, the polar form of z is -âˆš2(cos Î±/2 – sin Î±/2){cos (Î±/2 – sin 3Ï€/4) + i sin (Î±/2 – sin 3Ï€/4)}..

Case 3: When 3Ï€/2 < Î± < 2Ï€

In this case, we have,

|z| = |âˆš2(cos Î±/2 – sin Î±/2)| = -âˆš2(cos Î±/2 – sin Î±/2)

And tan Î¸ = |tan (Ï€/4 + Î±/2)| = tan (Ï€/4 + Î±/2) = – tan {Ï€ – (Ï€/4 + Î±/2)} = tan (Î±/2 – 3Ï€/4)

Î¸ = Î±/2 – 3Ï€/4

Clearly, z lies in the first quadrant . Therefore, arg(z) = Î¸ = Î±/2 – 3Ï€/4

Therefore, the polar form of z is -âˆš2(cos Î±/2 – sin Î±/2){cos (Î±/2 – sin 3Ï€/4) + i sin (Î±/2 – sin 3Ï€/4)}.

(iv)

Let z =

Now, z =

= âˆš2

Let Î¸ be an acute angle given by tan Î¸ =

tan Î¸ =

tan Î¸ = |tan (Ï€/4 + Ï€/3)| = |tan 7Ï€/12|

Î¸ = 7Ï€/12

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

Therefore, the polar form of z is âˆš2(cos 7Ï€/12 – sin 7Ï€/12).

### Question 4. If z1 and z2 are two complex number such that |z1| = |z2| and arg (z1) + arg (z2) = Ï€, then show that .

Solution:

We are given |z1| = |z2| and arg (z1) + arg (z2) = Ï€. Suppose arg (z1) = Î¸, then arg (z2) = Ï€ â€“ Î¸.

We know z = |z| (cos Î¸ + i sin Î¸)

z2 = |z2| (cos (Ï€ â€“ Î¸) + i sin (Ï€ â€“ Î¸))

= |z2| (â€“cos Î¸ + i sin Î¸)

= â€“ |z2| (cos Î¸ â€“ i sin Î¸)

The conjugate of z2 = â€“ |z2| (cos Î¸ + i sin Î¸)

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

= |z2| (cos Î¸ + i sin Î¸)

= â€“ [â€“ |z2| (cos Î¸ + i sin Î¸)]

= 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,

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

= arg (z1) âˆ’ arg (z4) + arg (z2) âˆ’ arg (z3)

= [arg (z1) + arg (z2)] âˆ’ [arg (z3) + arg (z4)]

= 0 âˆ’ 0

= 0

= R.H.S

Hence proved.

### Question 6. Express sin Ï€/5 + i (1 â€“ cos Ï€/5) in polar form.

Solution:

We are given,

Z = sin Ï€/5 + i (1 â€“ cos Ï€/5)

= 2 sin Ï€/10 cos Ï€/10 + i (2 sin2 Ï€/10)

= 2 sin Ï€/10 (cos Ï€/10 + i sin Ï€/10)

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

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

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