RD Sharma Solutions for Class 11 Maths Chapter 13 Complex Numbers

RD Sharma Solutions Class 11 Maths Chapter 13 – Get Free PDF (for 2021 – 2022)

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers are provided here for students to prepare and score good marks in the board exams. The combination of a real number and an imaginary number is termed as a complex number. All the concepts related to complex numbers are explained in this chapter with suitable examples.

The RD Sharma Solutions for Class 11 Maths are formulated by experts at BYJU’S after conducting vast research on each concept. Students are provided with exercise-wise solutions to help understand the concepts clearly from the exam point of view. Experts have prepared the solutions where the concepts are explained in detail, which is very helpful for preparing for their board exams. Students who find difficulty in solving problems can quickly jump to RD Sharma Class 11 Maths Solutions, download the pdf from the links given below and can start practising offline for good results.

Chapter 13 – Complex Numbers contains four exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Integral Powers of IOTA (i).
  • Imaginary quantities.
  • Complex numbers.
  • Equality of complex numbers.
  • Addition of complex numbers.
    • Properties of addition of complex numbers.
  • Subtraction of complex numbers.
  • Multiplication of complex numbers.
    • Properties of multiplication.
  • Division of complex numbers.
  • Conjugate of complex numbers.
  • Modulus of a complex number.
    • Properties of modulus.
  • Reciprocal of a complex number.
  • Square roots of a complex number.
  • Representation of a complex number.
    • Geometrical representation of a complex number.
    • Argument or amplitude of a complex number for different signs of real and imaginary parts.
    • Vectorial representation of a complex number.
    • Polar or trigonometrical form of a complex number.
    • Multiplication of a complex number by IOTA.
    • The polar form of a complex number for different signs of real and imaginary parts.

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Access answers to RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers

EXERCISE 13.1 PAGE NO: 13.3

1. Evaluate the following:

(i) i 457

(ii) i 528

(iii) 1/ i58

(iv) i 37 + 1/i 67

(v) [i 41 + 1/i 257]

(vi) (i 77 + i 70 + i 87 + i 414)3

(vii) i 30 + i 40 + i 60

(viii) i 49 + i 68 + i 89 + i 110

Solution:

(i) i 457

Let us simplify we get,

i457 = i (456 + 1)

= i 4(114) × i

= (1)114 × i

= i [since i4 = 1]

(ii) i 528

Let us simplify we get,

i 528 = i 4(132)

= (1)132

= 1 [since i4 = 1]

(iii) 1/ i58

Let us simplify we get,

1/ i58 = 1/ i 56+2

= 1/ i 56 × i2

= 1/ (i4)14 × i2

= 1/ i2 [since, i4 = 1]

= 1/-1 [since, i2 = -1]

= -1

(iv) i 37 + 1/i 67

Let us simplify we get,

i 37 + 1/i 67 = i36+1 + 1/ i64+3

= i + 1/i3 [since, i4 = 1]

= i + i/i4

= i + i

= 2i

(v) [i 41 + 1/i 257]

Let us simplify we get,

[i 41 + 1/i 257] = [i40+1 + 1/ i256+1]

= [i + 1/i]9 [since, 1/i = -1]

= [i – i]

= 0

(vi) (i 77 + i 70 + i 87 + i 414)3

Let us simplify we get,

(i 77 + i 70 + i 87 + i 414)3 = (i(76 + 1) + i(68 + 2) + i(84 + 3) + i(412 + 2) ) 3

= (i + i2 + i3 + i2)3 [since i3 = – i, i2 = – 1]

= (i + (– 1) + (– i) + (– 1)) 3

= (– 2)3

= – 8

(vii) i 30 + i 40 + i 60

Let us simplify we get,

i 30 + i 40 + i 60 = i(28 + 2) + i40 + i60

= (i4)7 i2 + (i4)10 + (i4)15

= i2 + 110 + 115

= – 1 + 1 + 1

= 1

(viii) i 49 + i 68 + i 89 + i 110

Let us simplify we get,

i 49 + i 68 + i 89 + i 110 = i(48 + 1) + i68 + i(88 + 1) + i(116 + 2)

= (i4)12×i + (i4)17 + (i4)22×i + (i4)29×i2

= i + 1 + i – 1

= 2i

2. Show that 1 + i10 + i20 + i30 is a real number?

Solution:

Given:

1 + i10 + i20 + i30 = 1 + i(8 + 2) + i20 + i(28 + 2)

= 1 + (i4)2 × i2 + (i4)5 + (i4)7 × i2

= 1 – 1 + 1 – 1 [since, i4 = 1, i2 = – 1]

= 0

Hence , 1 + i10 + i20 + i30 is a real number.

3. Find the values of the following expressions:

(i) i49 + i68 + i89 + i110

(ii) i30 + i80 + i120

(iii) i + i2 + i3 + i4

(iv) i5 + i10 + i15

(v) [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574]

(vi) 1 + i2 + i4 + i6 + i8 + … + i20

(vii) (1 + i)6 + (1 – i)3

Solution:

(i) i49 + i68 + i89 + i110

Let us simplify we get,

i49 + i68 + i89 + i110 = i (48 + 1) + i68 + i(88 + 1) + i(108 + 2)

= (i4)12 × i + (i4)17 + (i4)22 × i + (i4)27 × i2

= i + 1 + i – 1 [since i4 = 1, i2 = – 1]

= 2i

∴ i49 + i68 + i89 + i110 = 2i

(ii) i30 + i80 + i120

Let us simplify we get,

i30 + i80 + i120 = i(28 + 2) + i80 + i120

= (i4)7 × i2 + (i4)20 + (i4)30

= – 1 + 1 + 1 [since i4 = 1, i2 = – 1]

= 1

∴ i30 + i80 + i120 = 1

(iii) i + i2 + i3 + i4

Let us simplify we get,

i + i2 + i3 + i4 = i + i2 + i2×i + i4

= i – 1 + (– 1) × i + 1 [since i4 = 1, i2 = – 1]

= i – 1 – i + 1

= 0

∴ i + i2 + i3 + i4 = 0

(iv) i5 + i10 + i15

Let us simplify we get,

i5 + i10 + i15 = i(4 + 1) + i(8 + 2) + i(12 + 3)

= (i4)1×i + (i4)2×i2 + (i4)3×i3

= (i4)1×i + (i4)2×i2 + (i4)3×i2×i

= 1×i + 1 × (– 1) + 1 × (– 1)×i

= i – 1 – i

= – 1

∴ i5 + i10 + i15 = -1

(v) [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574]

Let us simplify we get,

[i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574]

= [i10 (i582 + i580 + i578 + i576 + i574) / (i582 + i580 + i578 + i576 + i574)]

= i10

= i8 i2

= (i4)2 i2

= (1)2 (-1) [since i4 = 1, i2 = -1]

= -1 

∴ [i592 + i590 + i588 + i586 + i584] / [i582 + i580 + i578 + i576 + i574] = -1

(vi) 1 + i2 + i4 + i6 + i8 + … + i20

Let us simplify we get,

1 + i2 + i4 + i6 + i8 + … + i20 = 1 + (– 1) + 1 + (– 1) + 1 + … + 1

= 1

∴ 1 + i2 + i4 + i6 + i8 + … + i20 = 1

(vii) (1 + i)6 + (1 – i)3

Let us simplify we get,

(1 + i)6 + (1 – i)3 = {(1 + i)2 }3 + (1 – i)2 (1 – i)

= {1 + i2 + 2i}3 + (1 + i2 – 2i)(1 – i)

= {1 – 1 + 2i}3 + (1 – 1 – 2i)(1 – i)

= (2i)3 + (– 2i)(1 – i)

= 8i3 + (– 2i) + 2i2

= – 8i – 2i – 2 [since i3 = – i, i2 = – 1]

= – 10 i – 2

= – 2(1 + 5i)

= – 2 – 10i

∴ (1 + i)6 + (1 – i)3 = – 2 – 10i


EXERCISE 13.2 PAGE NO: 13.31

1. Express the following complex numbers in the standard form a + ib:

(i) (1 + i) (1 + 2i)

(ii) (3 + 2i) / (-2 + i)

(iii) 1/(2 + i)2

(iv) (1 – i) / (1 + i)

(v) (2 + i)3 / (2 + 3i)

(vi) [(1 + i) (1 +√3i)] / (1 – i)

(vii) (2 + 3i) / (4 + 5i)

(viii) (1 – i)3 / (1 – i3)

(ix) (1 + 2i)-3

(x) (3 – 4i) / [(4 – 2i) (1 + i)]

(xi)

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 1

(xii) (5 +√2i) / (1-√2i)

Solution:

(i) (1 + i) (1 + 2i)

Let us simplify and express in the standard form of (a + ib),

(1 + i) (1 + 2i) = (1+i)(1+2i)

= 1(1+2i)+i(1+2i)

= 1+2i+i+2i2

= 1+3i+2(-1) [since, i2 = -1]

= 1+3i-2

= -1+3i

∴ The values of a, b are -1, 3.

(ii) (3 + 2i) / (-2 + i)

Let us simplify and express in the standard form of (a + ib),

(3 + 2i) / (-2 + i) = [(3 + 2i) / (-2 + i)] × (-2-i) / (-2-i) [multiply and divide with (-2-i)]

= [3(-2-i) + 2i (-2-i)] / [(-2)2 – (i)2]

= [-6 -3i – 4i -2i2] / (4-i2)

= [-6 -7i -2(-1)] / (4 – (-1)) [since, i2 = -1]

= [-4 -7i] / 5

∴ The values of a, b are -4/5, -7/5

(iii) 1/(2 + i)2

Let us simplify and express in the standard form of (a + ib),

1/(2 + i)2 = 1/(22 + i2 + 2(2) (i))

= 1/ (4 – 1 + 4i) [since, i2 = -1]

= 1/(3 + 4i) [multiply and divide with (3 – 4i)]

= 1/(3 + 4i) × (3 – 4i)/ (3 – 4i)]

= (3-4i)/ (32 – (4i)2)

= (3-4i)/ (9 – 16i2)

= (3-4i)/ (9 – 16(-1)) [since, i2 = -1]

= (3-4i)/25

∴ The values of a, b are 3/25, -4/25

(iv) (1 – i) / (1 + i)

Let us simplify and express in the standard form of (a + ib),

(1 – i) / (1 + i) = (1 – i) / (1 + i) × (1-i)/(1-i) [multiply and divide with (1-i)]

= (12 + i2 – 2(1)(i)) / (12 – i2)

= (1 + (-1) -2i) / (1 – (-1))

= -2i/2

= -i

∴ The values of a, b are 0, -1

(v) (2 + i)3 / (2 + 3i)

Let us simplify and express in the standard form of (a + ib),

(2 + i)3 / (2 + 3i) = (23 + i3 + 3(2)2(i) + 3(i)2(2)) / (2 + 3i)

= (8 + (i2.i) + 3(4)(i) + 6i2) / (2 + 3i)

= (8 + (-1)i + 12i + 6(-1)) / (2 + 3i)

= (2 + 11i) / (2 + 3i)

[multiply and divide with (2-3i)]

= (2 + 11i)/(2 + 3i) × (2-3i)/(2-3i)

= [2(2-3i) + 11i(2-3i)] / (22 – (3i)2)

= (4 – 6i + 22i – 33i2) / (4 – 9i2)

= (4 + 16i – 33(-1)) / (4 – 9(-1)) [since, i2 = -1]

= (37 + 16i) / 13

∴ The values of a, b are 37/13, 16/13

(vi) [(1 + i) (1 +√3i)] / (1 – i)

Let us simplify and express in the standard form of (a + ib),

[(1 + i) (1 +√3i)] / (1 – i) = [1(1+√3i) + i(1+√3i)] / (1-i)

= (1 + √3i + i + √3i2) / (1 – i)

= (1 + (√3+1)i + √3(-1)) / (1-i) [since, i2 = -1]

= [(1-√3) + (1+√3)i] / (1-i)

[multiply and divide with (1+i)]

= [(1-√3) + (1+√3)i] / (1-i) × (1+i)/(1+i)

= [(1-√3) (1+i) + (1+√3)i(1+i)] / (12 – i2)

= [1-√3+ (1-√3)i + (1+√3)i + (1+√3)i2] / (1-(-1)) [since, i2 = -1]

= [(1-√3)+(1-√3+1+√3)i+(1+√3)(-1)] / 2

= (-2√3 + 2i) / 2

= -√3 + i

∴ The values of a, b are -√3, 1

(vii) (2 + 3i) / (4 + 5i)

Let us simplify and express in the standard form of (a + ib),

(2 + 3i) / (4 + 5i) = [multiply and divide with (4-5i)]

= (2 + 3i) / (4 + 5i) × (4-5i)/(4-5i)

= [2(4-5i) + 3i(4-5i)] / (42 – (5i)2)

= [8 – 10i + 12i – 15i2] / (16 – 25i2)

= [8+2i-15(-1)] / (16 – 25(-1)) [since, i2 = -1]

= (23 + 2i) / 41

∴ The values of a, b are 23/41, 2/41

(viii) (1 – i)3 / (1 – i3)

Let us simplify and express in the standard form of (a + ib),

(1 – i)3 / (1 – i3) = [13 – 3(1)2i + 3(1)(i)2 – i3] / (1-i2.i)

= [1 – 3i + 3(-1)-i2.i] / (1 – (-1)i) [since, i2 = -1]

= [-2 – 3i – (-1)i] / (1+i)

= [-2-4i] / (1+i)

[Multiply and divide with (1-i)]

= [-2-4i] / (1+i) × (1-i)/(1-i)

= [-2(1-i)-4i(1-i)] / (12 – i2)

= [-2+2i-4i+4i2] / (1 – (-1))

= [-2-2i+4(-1)] /2

= (-6-2i)/2

= -3 – i

∴ The values of a, b are -3, -1

(ix) (1 + 2i)-3

Let us simplify and express in the standard form of (a + ib),

(1 + 2i)-3 = 1/(1 + 2i)3

= 1/(13+3(1)2 (2i)+2(1)(2i)2 + (2i)3)

= 1/(1+6i+4i2+8i3)

= 1/(1+6i+4(-1)+8i2.i) [since, i2 = -1]

= 1/(-3+6i+8(-1)i) [since, i2 = -1]

= 1/(-3-2i)

= -1/(3+2i)

[Multiply and divide with (3-2i)]

= -1/(3+2i) × (3-2i)/(3-2i)

= (-3+2i)/(32 – (2i)2)

= (-3+2i) / (9-4i2)

= (-3+2i) / (9-4(-1))

= (-3+2i) /13

∴ The values of a, b are -3/13, 2/13

(x) (3 – 4i) / [(4 – 2i) (1 + i)]

Let us simplify and express in the standard form of (a + ib),

(3 – 4i) / [(4 – 2i) (1 + i)] = (3-4i)/ [4(1+i)-2i(1+i)]

= (3-4i)/ [4+4i-2i-2i2]

= (3-4i)/ [4+2i-2(-1)] [since, i2 = -1]

= (3-4i)/ (6+2i)

[Multiply and divide with (6-2i)]

= (3-4i)/ (6+2i) × (6-2i)/(6-2i)

= [3(6-2i)-4i(6-2i)] / (62 – (2i)2)

= [18 – 6i – 24i + 8i2] / (36 – 4i2)

= [18 – 30i + 8 (-1)] / (36 – 4 (-1)) [since, i2 = -1]

= [10-30i] / 40

= (1 – 3i) / 4

∴ The values of a, b are 1/4, -3/4

(xi)

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 2
RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 3

(xii) (5 +√2i) / (1-√2i)

Let us simplify and express in the standard form of (a + ib),

(5 +√2i) / (1-√2i) = [Multiply and divide with (1+2i)]

= (5 +√2i) / (1-√2i) × (1+2i)/(1+2i)

= [5(1+2i) + 2i(1+2i)] / (12 – (2)2)

= [5+52i + 2i + 2i2] / (1 – 2i2)

= [5 + 62i + 2(-1)] / (1-2(-1)) [since, i2 = -1]

= [3+62i]/3

= 1+ 22i

∴ The values of a, b are 1, 22

2. Find the real values of x and y, if

(i) (x + iy) (2 – 3i) = 4 + i

(ii) (3x – 2i y) (2 + i)2 = 10(1 + i)

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 4

(iv) (1 + i) (x + iy) = 2 – 5i

Solution:

(i) (x + iy) (2 – 3i) = 4 + i

Given:

(x + iy) (2 – 3i) = 4 + i

Let us simplify the expression we get,

x(2 – 3i) + iy(2 – 3i) = 4 + i

2x – 3xi + 2yi – 3yi2 = 4 + i

2x + (-3x+2y)i – 3y (-1) = 4 + i [since, i2 = -1]

2x + (-3x+2y)i + 3y = 4 + i [since, i2 = -1]

(2x+3y) + i(-3x+2y) = 4 + i

Equating Real and Imaginary parts on both sides, we get

2x+3y = 4… (i)

And -3x+2y = 1… (ii)

Multiply (i) by 3 and (ii) by 2 and add

On solving we get,

6x – 6x – 9y + 4y = 12 + 2

13y = 14

y = 14/13

Substitute the value of y in (i) we get,

2x+3y = 4

2x + 3(14/13) = 4

2x = 4 – (42/13)

= (52-42)/13

2x = 10/13

x = 5/13

x = 5/13, y = 14/13

∴ The real values of x and y are 5/13, 14/13

(ii) (3x – 2i y) (2 + i)2 = 10(1 + i)

Given:

(3x – 2iy) (2+i)2 = 10(1+i)

(3x – 2yi) (22+i2+2(2)(i)) = 10+10i

(3x – 2yi) (4 + (-1)+4i) = 10+10i [since, i2 = -1]

(3x – 2yi) (3+4i) = 10+10i

Let us divide with 3+4i on both sides we get,

(3x – 2yi) = (10+10i)/(3+4i)

= Now multiply and divide with (3-4i)

= [10(3-4i) + 10i(3-4i)] / (32 – (4i)2)

= [30-40i+30i-40i2] / (9 – 16i2)

= [30-10i-40(-1)] / (9-16(-1))

= [70-10i]/25

Now, equating Real and Imaginary parts on both sides we get

3x = 70/25 and -2y = -10/25

x = 70/75 and y = 1/5

x = 14/15 and y = 1/5

∴ The real values of x and y are 14/15, 1/5

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 5

(4+2i) x-3i-3 + (9-7i)y = 10i

(4x+9y-3) + i(2x-7y-3) = 10i

Now, equating Real and Imaginary parts on both sides we get,

4x+9y-3 = 0 … (i)

And 2x-7y-3 = 10

2x-7y = 13 … (ii)

Multiply (i) by 7 and (ii) by 9 and add

On solving these equations we get

28x + 18x + 63y – 63y = 117 + 21

46x = 117 + 21

46x = 138

x = 138/46

= 3

Substitute the value of x in (i) we get,

4x+9y-3 = 0

9y = -9

y = -9/9

= -1

x = 3 and y = -1

∴ The real values of x and y are 3 and -1

(iv) (1 + i) (x + iy) = 2 – 5i

Given:

(1 + i) (x + iy) = 2 – 5i

Divide with (1+i) on both the sides we get,

(x + iy) = (2 – 5i)/(1+i)

Multiply and divide by (1-i)

= (2 – 5i)/(1+i) × (1-i)/(1-i)

= [2(1-i) – 5i (1-i)] / (12 – i2)

= [2 – 7i + 5(-1)] / 2 [since, i2 = -1]

= (-3-7i)/2

Now, equating Real and Imaginary parts on both sides we get

x = -3/2 and y = -7/2

∴ Thee real values of x and y are -3/2, -7/2

3. Find the conjugates of the following complex numbers:

(i) 4 – 5i

(ii) 1 / (3 + 5i)

(iii) 1 / (1 + i)

(iv) (3 – i)2 / (2 + i)

(v) [(1 + i) (2 + i)] / (3 + i)

(vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)]

Solution:

(i) 4 – 5i

Given:

4 – 5i

We know the conjugate of a complex number (a + ib) is (a – ib)

So,

∴ The conjugate of (4 – 5i) is (4 + 5i)

(ii) 1 / (3 + 5i)

Given:

1 / (3 + 5i)

Since the given complex number is not in the standard form of (a + ib)

Let us convert to standard form by multiplying and dividing with (3 – 5i)

We get,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 6

We know the conjugate of a complex number (a + ib) is (a – ib)

So,

∴ The conjugate of (3 – 5i)/34 is (3 + 5i)/34

(iii) 1 / (1 + i)

Given:

1 / (1 + i)

Since the given complex number is not in the standard form of (a + ib)

Let us convert to standard form by multiplying and dividing with (1 – i)

We get,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 7

We know the conjugate of a complex number (a + ib) is (a – ib)

So,

∴ The conjugate of (1-i)/2 is (1+i)/2

(iv) (3 – i)2 / (2 + i)

Given:

(3 – i)2 / (2 + i)

Since the given complex number is not in the standard form of (a + ib)

Let us convert to standard form,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 8

We know the conjugate of a complex number (a + ib) is (a – ib)

So,

∴ The conjugate of (2 – 4i) is (2 + 4i)

(v) [(1 + i) (2 + i)] / (3 + i)

Given:

[(1 + i) (2 + i)] / (3 + i)

Since the given complex number is not in the standard form of (a + ib)

Let us convert to standard form,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 9

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 10

We know the conjugate of a complex number (a + ib) is (a – ib)

So,

∴ The conjugate of (3 + 4i)/5 is (3 – 4i)/5

(vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)]

Given:

[(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)]

Since the given complex number is not in the standard form of (a + ib)

Let us convert to standard form,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 11

We know the conjugate of a complex number (a + ib) is (a – ib)

So,

∴ The conjugate of (63 – 16i)/25 is (63 + 16i)/25

4. Find the multiplicative inverse of the following complex numbers:

(i) 1 – i

(ii) (1 + i √3)2

(iii) 4 – 3i

(iv) √5 + 3i

Solution:

(i) 1 – i

Given:

1 – i

We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z

So,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 12

∴ The multiplicative inverse of (1 – i) is (1 + i)/2

(ii) (1 + i √3)2

Given:

(1 + i √3)2

Z = (1 + i √3)2

= 12 + (i √3)2 + 2 (1) (i√3)

= 1 + 3i2 + 2 i√3

= 1 + 3(-1) + 2 i√3 [since, i2 = -1]

= 1 – 3 + 2 i√3

= -2 + 2 i√3

We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z

So,

Z = -2 + 2 i√3

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 13

∴ The multiplicative inverse of (1 + i√3)2 is (-1-i√3)/8

(iii) 4 – 3i

Given:

4 – 3i

We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z

So,

Z = 4 – 3i

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 14

∴ The multiplicative inverse of (4 – 3i) is (4 + 3i)/25

(iv) √5 + 3i

Given:

√5 + 3i

We know the multiplicative inverse of a complex number (Z) is Z-1 or 1/Z

So,

Z = √5 + 3i

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 15

∴ The multiplicative inverse of (√5 + 3i) is (√5 – 3i)/14

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 16

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 17

6. If z1 = (2 – i), z2 = (-2 + i), find

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 18

Solution:

Given:

z1 = (2 – i) and z2 = (-2 + i)

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 19

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 19

7. Find the modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)]

Solution:

Given:

[(1 + i)/(1 – i)] – [(1 – i)/(1 + i)]

So,

Z = [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)]

Let us simplify, we get

= [(1+i) (1+i) – (1-i) (1-i)] / (12 – i2)

= [12 + i2 + 2(1)(i) – (12 + i2 – 2(1)(i))] / (1 – (-1)) [Since, i2 = -1]

= 4i/2

= 2i

We know that for a complex number Z = (a+ib) it’s magnitude is given by |z| = (a2 + b2)

So,

|Z| = (02 + 22)

= 2

∴ The modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)] is 2.

8. If x + iy = (a+ib)/(a-ib), prove that x2 + y2 = 1

Solution:

Given:

x + iy = (a+ib)/(a-ib)

We know that for a complex number Z = (a+ib) it’s magnitude is given by |z| = (a2 + b2)

So,

|a/b| is |a| / |b|

Applying Modulus on both sides we get,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 21

9. Find the least positive integral value of n for which [(1+i)/(1-i)]n is real.

Solution:

Given:

[(1+i)/(1-i)]n

Z = [(1+i)/(1-i)]n

Now let us multiply and divide by (1+i), we get

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 22

= i [which is not real]

For n = 2, we have

[(1+i)/(1-i)]2 = i2

= -1 [which is real]

So, the smallest positive integral ‘n’ that can make [(1+i)/(1-i)]n real is 2.

∴ The smallest positive integral value of ‘n’ is 2.

10. Find the real values of θ for which the complex number (1 + i cos θ) / (1 – 2i cos θ) is purely real.

Solution:

Given:

(1 + i cos θ) / (1 – 2i cos θ)

Z = (1 + i cos θ) / (1 – 2i cos θ)

Let us multiply and divide by (1 + 2i cos θ)

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 23

For a complex number to be purely real, the imaginary part should be equal to zero.

So,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 24

3cos θ = 0 (since, 1 + 4cos2θ ≥ 1)

cos θ = 0

cos θ = cos π/2

θ = [(2n+1)π] / 2, for n ∈ Z

= 2nπ ± π/2, for n ∈ Z

∴ The values of θ to get the complex number to be purely real is 2nπ ± π/2, for n ∈ Z

11. Find the smallest positive integer value of n for which (1+i) n / (1-i) n-2 is a real number.

Solution:

Given:

(1+i) n / (1-i) n-2

Z = (1+i) n / (1-i) n-2

Let us multiply and divide by (1 – i)2

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 25

For n = 1,

Z = -2i1+1

= -2i2

= 2, which is a real number.

∴ The smallest positive integer value of n is 1.

12. If [(1+i)/(1-i)]3 – [(1-i)/(1+i)]3 = x + iy, find (x, y)

Solution:

Given:

[(1+i)/(1-i)]3 – [(1-i)/(1+i)]3 = x + iy

Let us rationalize the denominator, we get

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 26

i3–(-i)3 = x + iy

2i3 = x + iy

2i2.i = x + iy

2(-1)I = x + iy

-2i = x + iy

Equating Real and Imaginary parts on both sides we get

x = 0 and y = -2

∴ The values of x and y are 0 and -2.

13. If (1+i)2 / (2-i) = x + iy, find x + y

Solution:

Given:

(1+i)2 / (2-i) = x + iy

Upon expansion we get,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 27

Let us equate real and imaginary parts on both sides we get,

x = -2/5 and y = 4/5

so,

x + y = -2/5 + 4/5

= (-2+4)/5

= 2/5

∴ The value of (x + y) is 2/5


EXERCISE 13.3 PAGE NO: 13.39

1. Find the square root of the following complex numbers.

(i) – 5 + 12i

(ii) -7 – 24i

(iii) 1 – i

(iv) – 8 – 6i

(v) 8 – 15i

(vi) -11 – 60√-1

(vii) 1 + 4√-3

(viii) 4i

(ix) -i

Solution:

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 28

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 29

(i) – 5 + 12i

Given:

– 5 + 12i

We know, Z = a + ib

So, (a + ib) = (-5+12i)

Here, b > 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 30
RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 31

∴ Square root of (– 5 + 12i) is ±[2 + 3i]

(ii) -7 – 24i

Given:

-7 – 24i

We know, Z = -7 – 24i

So, (a + ib) = √(-7 – 24i)

Here, b < 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 32

∴ Square root of (-7 – 24i) is ± [3 – 4i]

(iii) 1 – i

Given:

1 – i

We know, Z = (1 – i)

So, (a + ib) = (1 – i)

Here, b < 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 33

∴ Square root of (1 – i) is ± [((2+1)/2) – i ((2-1)/2)]

(iv) -8 -6i

Given:

-8 -6i

We know, Z = -8 -6i

So, (a + ib) = -8 -6i

Here, b < 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 34

= [11/2 – i 91/2]

= ± [1 – 3i]

∴ Square root of (-8 -6i) is ± [1 – 3i]

(v) 8 – 15i

Given:

8 – 15i

We know, Z = 8 – 15i

So, (a + ib) = 8 – 15i

Here, b < 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 35

∴ Square root of (8 – 15i) is ± 1/2 (5 – 3i)

(vi) -11 – 60-1

Given:

-11 – 60-1

We know, Z = -11 – 60-1

So, (a + ib) = -11 – 60-1

= -11 – 60i

Here, b < 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 36

∴ Square root of (-11 – 60-1) is ± (5 – 6i)

(vii) 1 + 4-3

Given:

1 + 4-3

We know, Z = 1 + 4-3

So, (a + ib) = 1 + 4-3

= 1 + 4(3) (-1)

= 1 + 43i

Here, b > 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 37
RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 38

∴ Square root of (1 + 4-3) is ± (2 + 3i)

(viii) 4i

Given:

4i

We know, Z = 4i

So, (a + ib) = 4i

Here, b > 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 39

∴ Square root of 4i is ± 2 (1 + i)

(ix) –i

Given:

-i

We know, Z = -i

So, (a + ib) = -i

Here, b < 0

Let us simplify now,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 40

∴ Square root of –i is ± 1/2 (1 – i)


EXERCISE 13.4 PAGE NO: 13.57

1. Find the modulus and arguments 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)

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 1st 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 1st 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 4th 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))

∴ 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

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 41

= 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 4th 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 (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

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 42

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 4th 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

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 43

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 2nd 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 120o – i cos 120o

Given: Z = sin 120o – i cos 120o

= 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 1st 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 – i3), we get

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 44

So now,

|Z| = (x2 + y2)

= ((-4)2 + (43)2)

= (16 + 48)

= (64)

= 8

θ = tan-1 (|y| / |x|)

= tan-1 ((43) / 4)

= tan-1 (3)

Since x < 0, y > 0 complex number lies in 2nd 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 4th 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 4th 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 4th 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 3th 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,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 46

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,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 47

∴ 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,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 48

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

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 49

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 – i3), we get

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 50

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,

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 51

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 52

Since x < 0, y < 0 complex number lies in 3rd 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))

4. If z1 and z2 are two complex number such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 53

Solution:

Given:

|z1| = |z2| and arg (z1) + arg (z2) = π

Let us assume arg (z1) = θ

arg (z2) = π – θ

We know that in the polar form, z = |z| (cos θ + i sin θ)

z1 = |z1| (cos θ + i sin θ) …………. (i)

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

= |z2| (-cos θ + i sin θ)

= – |z2| (cos θ – i sin θ)

Now let us find the conjugate of

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 54= – |z2| (cos θ + i sin θ) …… (ii) (since, \(|\overline{{Z}_{2}}|=|Z_{2}|\))

Now,

z1 /
RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 56= [|z1| (cos θ + i sin θ)] / [-|z2| (cos θ + i sin θ)]

= – |z1| / |z2| [since, |z1| = |z2|]

= -1

When we cross multiply we get,

z1 = –
RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 57

Hence proved.

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

Solution:

Given:

RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers image - 58

Hence proved.

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

Solution:

Given:

Z = sin π/5 + i (1 – cos π/5)

By using the formula,

sin 2θ = 2 sin θ cos θ

1- cos 2θ = 2 sin2 θ

So,

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

= 2 sin π/10 (cos π/10 + i sin π/10)

∴ The polar form of sin π/5 + i (1 – cos π/5) is 2 sin π/10 (cos π/10 + i sin π/10)


Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 13 – Complex Numbers

Exercise 13.1 Solutions

Exercise 13.2 Solutions

Exercise 13.3 Solutions

Exercise 13.4 Solutions

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